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Mathematics - Dissertations
Mathematics
8-2012
The Clar Structure of Fullerenes
Elizabeth Jane Hartung
Syracuse University
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Hartung, Elizabeth Jane, "The Clar Structure of Fullerenes" (2012). Mathematics - Dissertations. Paper 69.
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ABSTRACT
A fullerene is a 3-regular plane graph Γ = (V, E, F ) consisting only of
pentagonal and hexagonal faces. Fullerenes are designed to model carbon
molecules. The Clar number and Fries number are two parameters that
are related to the stability of carbon molecules. We introduce chain decompositions, a new method to find lower bounds for the Clar and Fries
numbers. In Chapter 3, we define the Clar structure for a fullerene, a less
general decomposition designed to compute the Clar number for classes of
fullerenes. We use these new decompositions to understand the structure
of fullerenes and achieve several results. In Chapter 4, we classify and give
a construction for all fullerenes on |V | vertices that attain the maximum
Clar number
|V |
6
− 2. In Chapter 5, we settle an open question with a
counterexample: we construct an infinite family of fullerenes for which a
set of faces attaining the Clar number cannot be a subset of a set of faces
that attains the Fries number. We develop a method to calculate the Clar
number directly for many infinite families of fullerenes.
THE CLAR STRUCTURE OF FULLERENES
by
Elizabeth J. Hartung
B.S., Indiana University of Pennsylvania, 2006
M.S., Syracuse University, 2008
Dissertation
Submitted in partial fulfillment of the requirements for the degree of
Doctor of Philosophy in Mathematics.
Syracuse University
August 2012
c Elizabeth J. Hartung 2012
Copyright All Rights Reserved
iv
Acknowledgments
I would like to thank my advisor, Jack Graver, for his mentorship and friendship.
Throughout this process, he has been motivating, patient, and inspiring, and I feel
incredibly fortunate to work with someone I respect so deeply. I am grateful for the
encouragement and support that I have received from my family. In particular, I
would like to thank my parents, Paul and Linda Hartung. I could not have done
without the reassurance, community, and perspective provided by friends, especially
Benjamin, Katie, Lisa, Tony, and Kanat. Finally, I would like to thank the members
of my committee for their time and thoughtful consideration.
v
Contents
Acknowledgments
1 Introduction
iv
1
1.1
Background and Definitions . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
The Clar, Fries, and Face-Independence Numbers . . . . . . . . . . .
3
2 Chains and Improper Face 3-Colorings
16
2.1
Improper Face 3-Colorings . . . . . . . . . . . . . . . . . . . . . . . .
16
2.2
Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
3 Clar Structures
34
3.1
Clar Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34
3.2
Calculating the Number of Edges in A over a Clar Chain . . . . . . .
49
4 Fullerenes with Complete Clar Structure
60
4.1
The Leapfrog Construction . . . . . . . . . . . . . . . . . . . . . . . .
61
4.2
Fullerenes with Complete Clar Structures . . . . . . . . . . . . . . . .
64
CONTENTS
vi
5 Clar and Fries Class
75
5.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
5.2
Basic Patches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
77
5.2.1
A choice for the void and Clar faces that requires two Kekulé
extensions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.2
Extending Kekulé structures over basic patches with other choices
for the void and Clar faces . . . . . . . . . . . . . . . . . . . .
5.3
77
79
Fullerenes over which the Clar and Fries numbers cannot be attained
simultaneously
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 Examples and Future Research
82
86
6.1
Two Classes of Widely Separated Fullerenes . . . . . . . . . . . . . .
87
6.2
Icosahedral Leapfrog Fullerenes . . . . . . . . . . . . . . . . . . . . .
92
6.3
Future Research . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
96
vii
List of Figures
1.1
Two Different Kekulé Structures. The thick blue edges represent edges
in the Kekulé structure. . . . . . . . . . . . . . . . . . . . . . . . . .
1.2
Kekulé structure with face 3-coloring. The void faces are in the blue
color class. Thick blue edges represent edges in the Kekulé structure.
1.3
4
5
A clear field between two pentagons drawn in the hexagonal tessellation. The pentagonal faces are blue. At each pentagon, a 60◦ wedge is
cut out and the two rays bounding the wedge are identified. . . . . .
1.4
9
Pairs of pentagons with Coxeter coordinates (4,1) and Coxeter coordinate (4). The pentagons are shown in blue.
. . . . . . . . . . . . . .
10
1.5
The leapfrog construction on a patch of a plane graph, Ω. . . . . . . .
12
2.1
All possible non-crossing couplings over a face of Γ. The edges in K
2.2
are shown in red. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
Expansion of edges of distance 3 around f with face 3-coloring . . . .
24
LIST OF FIGURES
2.3
viii
An augmented chain f0 , t1 , f2 , ..., t8 , f8 between pentagons f0 and f8 .
Dark blue edges represent edges in T . . . . . . . . . . . . . . . . . . .
2.4
25
Case 2: The edges in T are dark blue; the faces in C are the yellow
faces not incident with edges in T . The yellow faces with green circles
are added to C to get C 0 . The large red vertices belong to V ∗ (C 0 ).
Over this chain, there are k = 16 edges in T and l = 5 non-consecutive
sharp turns. Thus the contribution to |V ∗ (C 0 )| is 4 × 16 − 6 × 5 = 34 .
2.5
29
The Kekulé scheme derived from D consists of edges joining two faces
in D. Dark blue edges represent Kekulé edges, red edges represent
edges in T . In each case, all vertices outside of the augmented chain
are covered by exactly one edge in the Kekulé scheme.
3.1
. . . . . . . .
A benzene chain. Edges in A exit yellow faces from adjacent vertices
to form a short closed chain. . . . . . . . . . . . . . . . . . . . . . . .
3.2
36
For Clar chains, edges in A coupled over pentagons must exit from
adjacent vertices. The indicated couplings are avoided. . . . . . . . .
3.3
31
38
A straight chain segment with six edges in A joining a pair of faces
with Coxeter coordinates (6, 6). Dark blue edges represent edges in A.
44
3.4
Chains can become complicated when more than two pentagons interact. 46
3.5
Pairs of pentagons connected by two straight chain segments. The
pentagonal faces are yellow. . . . . . . . . . . . . . . . . . . . . . . .
48
LIST OF FIGURES
3.6
ix
The Coxeter coordinates of the segment between pentagons p1 and p2
are (8, 2). If the yellow faces contain the set C, this chain is of Type 1.
If the red faces contain the set C, the chain is of Type 2. If the blue
faces contain the set C, this chain is of Type 3. . . . . . . . . . . . .
50
3.7
Chains of Type 1 and 2. Thick blue edges are edges in A. . . . . . . .
52
3.8
A chain of Type 3 begins with a pentagon p1 and must also start a
closed Clar chain through the hexagon h1 . Here the blue faces contain
the set C, the thick edges represent edges in A. . . . . . . . . . . . .
3.9
53
A chain of Type 3 connects pentagons p1 and p2 . Thick blue edges
represent edges in A. . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
4.1
Two pentagons paired by a single edge in A . . . . . . . . . . . . . .
61
4.2
The leapfrog transformation of a bipartite plane graph . . . . . . . .
63
4.3
Expansion of Γ over an edge in A . . . . . . . . . . . . . . . . . . . .
65
4.4
The reverse expansion around a quadrilateral face in (Θ` , P). . . . . .
67
4.5
Choose diagonal vertices for each quadrilateral in Θ. These vertices
correspond to opposite hexagons bounding a quadrilateral face in Θ` .
We collapse an edge of each of the paired hexagons to construct E −1 (Θ` , P). 70
4.6
If three quadrilaterals in Θ share a vertex, then any choice of diagonal
vertices for each quadrilateral includes some vertex twice. If at most
two quadrilaterals share each vertex, we can choose diagonal vertices
for each quadrilateral of Θ so that no vertex is chosen twice. . . . . .
72
LIST OF FIGURES
4.7
x
Two choices for contracting a quadrilateral face in Θ` . A different
choice is equivalent to a Stone-Wales transformation of the fullerene. .
5.1
The white faces represent a basic patch. The void faces are contained
in the set of blue faces, the Clar faces in the set of pink faces. . . . .
5.2
79
Over a basic patch, we choose other color classes for the void and Clar
faces and consider extensions of the resulting Kekulé structure. . . . .
5.4
77
Extension 1 and Extension 2 on a basic patch where the Clar faces are
pink and the void faces are blue. . . . . . . . . . . . . . . . . . . . . .
5.3
73
81
The vertices of this auxiliary graph represent the pentagons in a fullerene.
The edges give the Coxeter coordinates of the segments between nearby
pentagons.
5.5
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82
The Clar faces are red and the void faces are blue. An arrow indicates
the patch over which the Fries and Clar deficits cannot be minimized
simultaneously. Figure (a) minimizes the Clar deficit, Figure (b) minimizes the Fries deficit. The numerals represent faces in the sets B1
and B2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1
85
Class of Fullerenes that generalizes the family described in Chapter 5.
The pentagons are paired over green segments with Coxeter coordinates
(p, p).
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87
LIST OF FIGURES
6.2
xi
The auxiliary graph for a class of symmetric fullerenes. Vertices represent pentagons, edges show the Coxeter coordinates of segments between nearby pentagons. The pentagons paired over segments with
coordinates (r). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3
91
To form an icosahedral fullerene, replace each face of an icosahedron
with an equilateral triangle from the hexagonal tessellation with vertices at the centers of faces. . . . . . . . . . . . . . . . . . . . . . . .
93
6.4
Five equilateral triangles that share P3 as a vertex. . . . . . . . . . .
94
6.5
The Leapfrog Icosahedron with a pairing of pentagons, all of Type 1.
96
1
Chapter 1
Introduction
1.1
Background and Definitions
A Fullerene is a 3-regular plane graph Γ = (V, E, F ) consisting only of pentagonal and
hexagonal faces. Fullerenes are designed to model carbon molecules. The vertices of
the graph represent carbon atoms and edges represent chemical bonds between them.
The term “fullerene” is also used to denote the actual carbon molecule. A specific
fullerene with 60 atoms, C60 , was the first pure carbon molecule discovered and is
the most common naturally occurring carbon molecule. C60 was first synthesized by
Kroto, Heath, O’Brien, Curl and Smalley in 1985, and the discovery earned Kroto,
Curl, and Smalley a Nobel prize in 1996. In 1991, C60 was named Science Magazine’s
“Molecule of the Year.” Nanotubes are also examples of fullerenes. The smallest
fullerene is the dodecahedron.
CHAPTER 1. INTRODUCTION
2
Carbon atoms form four chemical bonds. Three of these are strong bonds, and
one is weak. In the graphical representation of a fullerene, we represent the three
strong bonds by the edges in the graph. The fourth bond is represented chemically
as a double bond.
A perfect matching of a graph is a set of edges of Γ such that each vertex is incident
with exactly one edge. Over a fullerene, the edges of a perfect matching correspond
to double bonds, and chemists call this matching a Kekulé structure. We refer to
edges in a given Kekulé structure as Kekulé edges. Petersen’s Theorem states that
in a bridgeless 3-regular graph, there is always a perfect matching [10]. We therefore
know that a fullerene always has at least one Kekulé structure. Propositions 1.1, 1.2,
and 1.4 are standard in the literature on fullerenes and are presented here for future
reference.
Proposition 1.1. Let K be a Kekulé structure on a fullerene Γ = (V, E, F ). Let P
be the set of pentagons and H be the set of hexagons in Γ. Then
1. |K| =
|V |
2
2. |E| =
3|V |
2
3. |P | = 12
4. |H| =
Proof.
|V |
2
− 10
CHAPTER 1. INTRODUCTION
3
1. Each vertex is incident with exactly one edge in the Kekulé structure, and every
edge in K is incident with two vertices. Thus |V | = 2|K|.
2. Every vertex is of degree 3, so 3|V | = 2|E|.
3. Summing together the number of vertices around each face gives 6|H| + 5|P |
vertices. Each vertex is incident with three faces, so 6|H| + 5|P | = 3|V |. Substituting 3|V | = 2|E| gives 2|E| = 6|H| + 5|P | and we know that |F | = |H| + |P |.
By Euler’s formula, |V | − |E| + |F | = 2, so 6|V | − 6|E| + 6|F | = 12, and by
substitution, 2(6|H| + 5|P |) − 3(6|H| + 5|P |) + 6(|H| + |P |) = 12. Simplifying
shows that |P | = 12.
4. 6|H| + 5|P | = 3|V | and |P | = 12, so 6|H| + 60 = 3|V |. Solving for |H| gives the
result.
1.2
The Clar, Fries, and Face-Independence Numbers
Given a Kekulé Structure on a fullerene Γ, a face of Γ may have 0, 1, 2 or 3 of its
bounding edges in K. The set of faces that have exactly i of their edges in K is
denoted by Bi (K). The faces in the set B0 (K) are called the void faces of K, and
the faces in the set B3 (K) are called the benzene faces of K. The number of benzene
CHAPTER 1. INTRODUCTION
(a) The red faces are benzene faces.
4
(b) In this Kekulé structure, none of the faces is
a benzene face.
Figure 1.1: Two Different Kekulé Structures. The thick blue edges represent edges
in the Kekulé structure.
faces over a fullerene is dependent upon which Kekulé structure is chosen. A patch of
a fullerene is pictured in Figure 1.1 with two different Kekulé structures. The thick
blue edges represent edges in each Kekulé structure. In Figure 1.1(a), the red faces
are all benzene faces; in Figure 1.1(b), none of the faces is a benzene face.
The Fries number of a fullerene Γ is the maximum number of benzene faces over
all possible Kekulé structures for Γ. We define a Fries set to be a set of benzene
faces that attains the Fries number for a Γ. The Clar number of a fullerene Γ is the
cardinality of a maximum independent set of benzene faces over all Kekulé structures
for Γ. We define a Clar set to be an independent set of benzene faces that attains the
Clar number for Γ. It has been found experimentally that the Clar and Fries numbers
for organic structures are related to the stability of the molecules. Clar and Gutman
CHAPTER 1. INTRODUCTION
5
Figure 1.2: Kekulé structure with face 3-coloring. The void faces are in the blue color
class. Thick blue edges represent edges in the Kekulé structure.
each theorized that this correlation would hold for carbon molecules [1].
It is natural to ask whether a Clar set is always a subset of some Fries set, or
equivalently, if there is a Kekulé structure that simultaneously gives a Fries set and
a Clar set. It was generally assumed that this was true. Settling this open question
was the initial research problem for this dissertation, and in Chapter 5 we show that
this assumption is false by constructing an infinite family of fullerenes for which a
Clar set is never a subset of a Fries set.
To motivate the next result we consider a hexagonal patch: a plane graph in
which all faces are hexagons except for possibly one outside face, all vertices are
of degree 2 or 3 and all vertices of degree 2 are incident with the outside face. A
hexagonal patch may be thought of as either a region of the hexagonal tessellation
of the plane or a region of a fullerene that includes no pentagons. As a region of the
CHAPTER 1. INTRODUCTION
6
hexagonal tessellation, a hexagonal patch inherits a face 3-coloring that is unique up
to interchanging the color classes. In a face 3-coloring of a hexagonal patch, each color
class is a maximal face-independent set. We construct a partial Kekulé structure over
such a patch in the following way: choose one color class to be the set of void faces.
Let all of the edges not bounding a void face be in the Kekuké structure. On the
interior of this patch, all of the faces in the other two color classes are benzene faces.
Figure 1.2 shows a patch with a face 3-coloring and associated Kekulé structure. The
blue faces are the void faces, the pink and yellow faces are all benzene faces and this
set is part of a potential Fries set. Furthermore, either the pink faces or the yellow
faces can be chosen as part of a potential Clar set. On the interior of this patch, twothirds of the faces are benzene faces and one-third of the faces form an independent
set of benzene faces. By Proposition 1.1 (3, 4), the number of faces in a fullerene is
approximately half the number of vertices. Thus we might expect the Fries number
of a fullerene Γ = (V, E, F ) to be bounded above by
bounded above by
|V |
.
6
|V |
3
and the Clar number to be
We see in the proposition below that this is indeed the case.
Proposition 1.2. Let K be a Kekulé structure for a fullerene Γ = (V, E, F ).
1. |B3 (K)| =
|V |
|B1 (K)| + 2|B2 (K)|
− δF (K), where δF (K) =
.
3
3
2. For an independent set X of faces in Γ, let V ∗ (X) be the set of vertices not
incident with a face in X and let P ∗ (X) be the set of pentagons not in the set
X. Then |X| =
|V |
|P ∗ (X)| + |V ∗ (X)|
+2−
.
6
6
CHAPTER 1. INTRODUCTION
7
3. For an independent set of faces X in B3 (K), |X| =
|V ∗ (X)|
.
δC (K, X) =
6
|V |
− δC (K, X) where
6
Proof.
1. Summing the number of edges from K bounding each face gives |B1 (K)| +
2|B2 (K)| + 3|B3 (K)| and counts each of the edges in K twice. Thus 2|K| =
|B1 (K)| + 2|B2 (K)| + 3|B3 (K)|, and by Proposition 1.1, |2K| = |V |. Solving
for |B3 (K)| gives |B3 (K)| =
|V | |B1 (K)| + 2|B2 (K)|
−
.
3
3
2. Since X is an independent set of faces, each vertex of Γ is incident with at
most one face in X. The total number of vertices incident with a face in X is 6
times the number of hexagons in X plus 5 times the number of pentagons in X:
6(|X|−(12−|P ∗ (X)|))+5(12−|P ∗ (X)|) = 6|X|−12+|P ∗ (X)| = |V |−|V ∗ (X)|.
Solving for |X| gives |X| =
|V |
6
+2−
|P ∗ (X)|+|V ∗ (X)|
.
6
3. Let X be an independent set of faces in B3 (K). Every benzene face must be a
hexagon, so P ∗ (X) = 12. Thus |X| =
|V |
6
−
|V ∗ (X)|
.
6
We call δF (K) the Fries deficit of K and δC (K, X) the Clar deficit of K with
respect to the set X. Fowler [3] and Graver [9] characterized fullerenes that attain
the maximum of
|V |
3
for the Fries number and gave a method for constructing all
fullerenes that attain this maximum. We show that the maximum possible Clar
CHAPTER 1. INTRODUCTION
number is
|V |
6
8
− 2 and in Chapter 4 we give a characterization and construction for
the fullerenes that achieve this maximum.
Over a region of a fullerene that includes a pentagon, a face 3-coloring is not possible and the Kekulé structure described above is disrupted. To compute or estimate
the Clar and Fries numbers, we isolate patches of the fullerene containing the pentagons so that the remainder of the fullerene consists solely of hexagons and admits
a face 3-coloring. One can then construct a partial Kekulé structure by selecting one
color class to be the void set as described above. The problem is then to extend the
partial Kekulé structure over the patches containing pentagons while minimizing the
Clar and Fries deficits. One such decomposition was introduced by Graver in [7], in
which pentagons were connected by a “spanning tree” of polygonal paths of hexagons.
In [7] and [8], Graver introduced clear fields and Coxeter coordinates, two concepts
that are used to describe the structure of fullerenes. Let Γ = (V, E, F ) be a fullerene
and let Γ∗ = (F, E, V ) be its dual. Two faces f1 and f2 are contained in a clear field
if all shortest dual paths between the corresponding vertices f1∗ and f2∗ consist only
of degree-6 vertices. We say that two faces of Γ are nearby if they are contained in a
clear field. A clear field between two pentagons is pictured in Figure 1.3 and we often
depict a patch of a fullerene containing pentagons in this way. We draw the patch in
the hexagonal tessellation of the plane. At each pentagon, we insert a 60◦ wedge and
identify the two rays bounding the wedge.
For a pair of nearby faces, the position of the two faces in relation to one another
CHAPTER 1. INTRODUCTION
9
Figure 1.3: A clear field between two pentagons drawn in the hexagonal tessellation.
The pentagonal faces are blue. At each pentagon, a 60◦ wedge is cut out and the two
rays bounding the wedge are identified.
is given by their Coxeter coordinates. Two faces are joined by edges running through
centers of hexagonal faces (corresponding to paths in the dual graph). A shortest
dual path between nearby faces can be oriented as two straight line segments with a
120◦ left turn between them, or in some cases, as one straight line segment. Examples
of these two cases are pictured in Figure 1.4. In the case with a 120◦ left turn, the
Coxeter coordinates are given as the ordered pair (m, n): a straight line segment
containing m faces before a 120◦ left turn, and another with n faces after the turn.
In the case with one straight line segment of length m, the coordinate is given just
CHAPTER 1. INTRODUCTION
10
Figure 1.4: Pairs of pentagons with Coxeter coordinates (4,1) and Coxeter coordinate
(4). The pentagons are shown in blue.
as (m).
One important family of fullerenes is the class of leapfrog fullerenes and their
construction is given in [4]. We define the leapfrog construction more generally for an
arbitrary plane graph. Given a plane graph Ω = (V, E, F ), the leapfrog construction
produces Ω` , a new 3-regular plane graph called the leapfrog graph of Ω. To construct
Ω` , first take the dual Ω∗ = (F, E, V ) and then take the snub of Ω∗ . For each vertex
f ∗ of degree j in Ω∗ , the snub construction replaces f ∗ with j new vertices, one for
each edge incident to f ∗ , and connects these j vertices in a j-cycle bounding a new
face f ` . (See Figure 1.5.) A face-only vertex covering of a plane graph is a set of faces
X such that each vertex is incident with exactly one face of X.
Lemma 1.3. Let Ω` be the leapfrog of the plane graph Ω = (V, E, F ). Ω` is 3-regular
and has a face corresponding to each face of Ω and a face corresponding to each vertex
of Ω. A face of Ω` corresponding to a face of degree j from Ω has degree j. A face
CHAPTER 1. INTRODUCTION
11
of Ω` corresponding to a vertex of degree k from Ω has degree 2k. The faces F ` of Ω`
corresponding to faces of Ω form a face-only vertex covering of Ω` .
Proof. A face f ∈ F of degree j becomes a vertex f ∗ of degree j in the dual Ω∗ . The
vertex f ∗ becomes a face of degree j when we take the snub of Ω∗ . A vertex v ∈ V
of degree k becomes a face v ∗ of degree of degree k in Ω∗ . When we take the snub
of Ω∗ , each vertex bounding the face v ∗ becomes an edge, and the corresponding face
fv in Ω` is of degree 2k. The snub operation replaces all vertices of Ω∗ by faces and
introduces new vertices only of degree 3: a vertex w of Ω` is incident with two edges
bounding a new face f ` (corresponding to a vertex f ∗ in Ω∗ ), and one edge that had
been incident to f ∗ in Ω∗ . We see that the vertex w is incident with f ` and that f `
is adjacent to only faces that correspond to vertices in Ω. Thus the faces in F ` form
a face-only vertex covering of Ω` .
Note that if Γ is a fullerene, then the leapfrog graph Γ` is also a fullerene: since
all vertices of Γ have degree 3, the faces of Γ` corresponding to these vertices are all
hexagons. All faces of Γ` corresponding to faces of Γ are hexagons and pentagons.
Fowler showed in [3] that a leapfrog fullerene on |V | vertices attains the Fries number
|V |
.
3
Graver proved in [9] that leapfrog fullerenes are the only fullerenes to satisfy
this upper bound. The face-independence number of a plane graph is the maximum
cardinality of a set of faces such that no two share an edge. By Proposition 1.2 the
face-independence number of a fullerene on |V | vertices is bounded above by
Combining the results from [3] and [9], we have the following proposition.
|V |
6
+ 2.
CHAPTER 1. INTRODUCTION
12
(a) A patch of a plane graph Ω.
(b) We first take the dual of Ω
(c) The dual of Ω, Ω∗
(d) We take the snub of Ω∗ .
(e) We now have the leapfrog graph Ω` .
Figure 1.5: The leapfrog construction on a patch of a plane graph, Ω.
CHAPTER 1. INTRODUCTION
13
Proposition 1.4. The following are equivalent for a fullerene Γ = (V, E, F ):
1. The Fries number of Γ is
|V |
;
3
2. The face-independence number of Γ is
|V |
6
+ 2;
3. Γ is a leapfrog fullerene;
4. For each pair of nearby pentagons with Coxeter coordinates (m, n), m ≡ n (mod
3) and for each pair of nearby pentagons with Coxeter coordinate (m), m ≡ 0
(mod 3).
The equivalence of (4) to the first three statements is not obvious; for a proof, see
Graver [9]. One may view Proposition 1.4 as a constructive characterization of those
fullerenes that achieve the maximum possible Fries number of
|V |
.
3
These leapfrog
fullerenes also achieve the maximum possible face independence number of
|V |
6
+ 2 by
including all pentagons in the independent set of faces. We show that the maximum
possible Clar number for a fullerene on |V | vertices is
|V |
6
− 2 and prove a theorem
parallel to Proposition 1.4 that constructively characterizes all fullerenes that achieve
the Clar number
|V |
6
− 2. The Clar number of a fullerene is less understood than the
Fries number, and the former parameter is the primary focus of this dissertation. For
small fullerenes, the Clar number has been calculated through computer searches.
We develop a theory to directly compute the Clar number for many infinite classes
of fullerenes.
In Chapter 2 we introduce a new decomposition designed to find lower bounds for
CHAPTER 1. INTRODUCTION
14
the Clar and Fries numbers. In Chapter 3, we create a less general decomposition
specifically conceived to compute the Clar number. Each of these decompositions is
based on chains, a new concept we introduce to describe the structure of fullerenes.
We define chains formally in the next chapter, but essentially chains are alternating
sequences of faces and edges that connect pairs of pentagons over a fullerene.
Consider a pentagon f0 and an edge e1 such that exactly one vertex of e1 is incident
with f0 and exactly one vertex of e1 is incident with a face f1 . If f1 is also a pentagon,
then f0 , e1 , f1 is the entire chain. If not, choose an edge e2 connecting f1 to a third face
f2 , and so on. In Chapter 2, we show that the twelve pentagons can be paired by six
chains such that the six chains have no edges in common. The structure of these chains
can be complicated. Chains can twist around one another, making computations for
calculation the Clar number difficult. Furthermore, there may be many different chain
decompositions and it is not clear which one(s) actually give the Clar number. For
our applications we consider a less general case. To avoid chains twisting around one
another, we introduce the concept of non-interfering chains, permitting us to connect
paired pentagons by chains of minimum length. To eliminate the possibility of still
better chain decompositions, we introduce the concept of widely separated pairs of
pentagons to ensure that all alternate chain decompositions have larger Clar deficits.
These conditions restrict the classes of fullerenes for which we can compute the Clar
number. However, this restriction allows us to achieve several major results: we
construct an infinite family of fullerenes for which no Clar set is contained in a Fries
CHAPTER 1. INTRODUCTION
15
set; we classify and give a construction for all fullerenes on |V | vertices that attain
the maximum Clar number
|V |
−2;
6
we develop a method to calculate the Clar number
directly for many infinite families of fullerenes.
16
Chapter 2
Chains and Improper Face
3-Colorings
2.1
Improper Face 3-Colorings
As noted in Chapter 1, the faces of a hexagonal patch of a fullerene may be 3-colored.
The following general theorem was proven by Saaty and Kainen. We state it here for
future reference.
Theorem 2.1 (Saaty and Kainen [11]). A 3-regular plane graph is face 3-colorable if
and only if each face has even degree.
Over a patch of faces that includes a pentagon, a proper face 3-coloring is clearly
not possible: the five faces adjacent to a pentagon cannot alternate in color. Thus
any face coloring of a fullerene using three colors is an improper face 3-coloring. If two
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
17
faces that share an edge have the same color, we call them incompatible faces. Let Γ
be a fullerene with a Kekulé structure K. We construct an improper face 3-coloring
based on this Kekulé structure. Given a face f of Γ, we say that an edge e exits f if
e shares exactly one vertex with f . We say that e lies on f if both vertices of e are
incident with f .
Lemma 2.2. Let Γ be a fullerene with Kekulé structure K.
1. An odd number of edges in K exit any pentagonal face of Γ. An even number
of edges (possibly 0) in K exit any hexagonal face.
2. Either all edges in K exiting f exit from consecutive vertices around f or f is
a hexagon with exactly two edges in K exiting f from opposite vertices.
Proof.
1. Let f be a face of Γ. If an edge in K lies on f , it is incident with two adjacent
vertices on f . Thus an even number of vertices (possibly 0) of f are covered by
edges in K that lie on f . If f is a hexagon, an even number of vertices remain
to be covered by edges from K that exit f . A pentagon always has an odd
number of vertices remaining from which edges in K exit.
2. Suppose exactly two edges e1 and e2 exit f . Then f is a hexagon, and either
e1 and e2 exit from vertices of f that are adjacent to one another or they are
separated by edges in K that lie on f , each of which covers two adjacent vertices.
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
18
Thus e1 and e2 exit from adjacent vertices or opposite vertices (of distance 3)
around f .
Suppose more than two edges in K exit a face f . We know that an odd number
of edges in K exit a pentagon and an even number of edges in K exit a hexagon.
If exactly three edges exit a pentagon or four edges exit a hexagon, then there
are two vertices on the face remaining to be covered by an edge in K that lies
on f , so these two vertices must be adjacent. Therefore, the edges from K that
exit f exit from consecutive vertices around f . If all vertices incident with f are
covered by edges that exit f , then these vertices clearly exit from consecutive
vertices around f .
For each face f with edges from K exiting f , we construct a coupling of these exit
edges. If an even number of edges in K exit f , we group the exit edges into pairs. If
an odd number of edges in K exit f , all but one of the exit edges are coupled over f .
For each pentagon, call the uncoupled exit edge the initial edge for that pentagon.
We say that the coupling of edges in K exiting f is non-crossing if we can connect
the exit edges through lines over f that do not cross one another (see Figure 2.1).
Lemma 2.3. For each face f of Γ, the edges in K that exit f admit a non-crossing
coupling. All possible couplings around a face are pictured in Figure 2.1.
Proof. Let h be a hexagon in Γ. By Lemma 2.2, an even number of edges in K exit
h. If there are exactly two exit edges, couple these edges. If more than two edges in
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
19
Figure 2.1: All possible non-crossing couplings over a face of Γ. The edges in K are
shown in red.
K exit h, these edges must be incident with consecutive vertices around h by Lemma
2.2. Couple these edges in the following way: Choose two edges in K that exit h
from adjacent vertices to form a couple. If there are two remaining edges in K that
exit h, couple these together. If there are four edges remaining to be coupled, couple
together two edges that exit from adjacent vertices of h. The remaining two edges
are then coupled and either exit from adjacent vertices of h or vertices on opposite
sides of h.
An odd number of edges in K exit each pentagon. If a single vertex of a pentagon
p is covered by an exit edge in K, this is the initial edge and is not coupled over
p. If more than one edge in K exits a pentagon, p, we know these edges exit from
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
20
consecutive vertices on p. Couple the edges in the following way: Choose one edge
to be the initial edge. If exactly two exit edges remain, couple them together. If four
remain, couple two consecutive exit edges, and let the remaining two be a couple.
In all cases, this is a non-crossing coupling. At most one pair of edges from K
exits from non-adjacent vertices around f . Two pairs of exit edges coupled over a
face f cross only if both pairs exit from vertices that are not adjacent.
Each edge in K that is not an initial edge from a pentagon has two couplings,
one over each face that it exits. The twelve initial edges have at most one coupling.
Given a coupling for a fullerene Γ, define a chain in Γ to be an alternating sequence
f0 , e1 , f1 , e2 , ..., ek , fk of faces fi of Γ and edges ei in K such that ei and ei+1 are
coupled edges exiting fi for 1 ≤ i ≤ k − 1, e1 exits f0 and ek exits fk . We require
further that if e1 (or ek ) is not an initial edge, f0 = fk and that e1 is coupled with ek
over f0 . If f0 = fk , we say that the chain is closed. A closed chain contains no initial
edges and creates a circuit. If the chain f0 , e1 , f1 , e2 , ..., ek , fk contains initial edges,
we say that the chain is open. In this case, e1 and ek are initial edges, so f0 and fk
are pentagons. A chain f0 , e1 , f1 , e2 , ..., ek , fk makes a sharp turn at fi if ei and ei+1
exit from adjacent vertices of fi .
The assigned coupling of exit edges over each face determines six open chains
between pairs of pentagons and ensures that each edge in K is included in exactly
one chain. By the construction of the coupling in Lemma 2.3, the chains may share
faces but do not cross one another. For the remainder of the chapter, we assume that
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
21
the couplings constructed are non-crossing.
Lemma 2.4. Let Γ be a fullerene with Kekulé structure K and a coupling assignment.
The set of edges in K decomposes into
1. Six open chains connecting pairs of pentagons;
2. Closed chains.
Proof. From the fullerene Γ = (V, E, F ) with a coupling, construct a new graph with
vertex set V and edge set K together with edges between coupled pairs. In this
graph, the twelve initial vertices exiting pentagons have degree 1 and the remaining
vertices have degree 2 (one edge from K and one from the coupling). Such a graph
decomposes into paths and circuits. The six paths that terminate at initial vertices of
pentagons correspond to the six open chains, and the remaining circuits correspond
to closed chains.
Let the set T denote the edges in K within the six open chains and S denote the
set of faces exited by edges from T . Ignoring the closed chains, we call the resulting
set of six open chains between pairs of pentagons a chain decomposition (S, T ) of Γ.
For a fullerene Γ with a chain decomposition (S, T ), we define an expansion E(S, T ) of
Γ as follows: “widen” each edge in T into a quadrilateral face by splitting its incident
vertices and joining each split pair by a new edge. Each vertex incident with an edge
in T becomes an edge of E(S, T ), each edge in T splits lengthwise into two edges of
E(S, T ).
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
22
Lemma 2.5. Let Γ be a fullerene with a chain decomposition (S, T ). The expansion
E(S, T ) of Γ is face 3-colorable.
Proof. The coupling determines the continuation of the chain, so either both edges of
a coupled pair are in T or neither edge is in T . Thus an even number of edges from
T exit a hexagon, and an odd number exit a pentagon. Each edge e in T that exits a
face f in S becomes a quadrilateral face of E(S, T ), and the vertex of e incident with
f becomes an edge. If a face f in S is exited by n edges from T , then its degree in the
expansion E(S, T ) is increased by n. Therefore, every face in the expansion E(S, T )
has even degree. E(S, T ) is now a 3-regular graph with all faces of even degree. Thus
by Theorem 2.1, E(S, T ) has a face 3-coloring.
Theorem 2.6. Let Γ be a fullerene with a chain decomposition (S, T ). Then there
is an improper face 3-coloring over which the incompatibilities occur exactly at faces
that share an edge in T . Up to a permutation of colors, this improper face 3-coloring
is unique.
Proof. Give the expansion E(S, T ) a proper face 3-coloring. We can return to the
fullerene Γ by collapsing each of the quadrilateral faces back into edges in T while
retaining the coloring of the remaining faces. Two faces of Γ that share an edge in T
correspond to opposite faces around a quadrilateral of E(S, T ), and accordingly have
the same color. Two faces of Γ that share an edge not in T share the same edge of
E(S, T ), hence they are assigned different colors.
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
23
Lemma 2.7. Let Γ be a fullerene with a chain decomposition (S, T ) and let f0 , t1 , f1 , ...tk , fk
be an open chain. In the associated improper face 3-coloring of Γ,
1. All faces f0 , f1 , ..., fk are in the same color class.
2. All faces incident with one of the edges t1 , t2 , ..., tk are in a second color class.
Proof. Consider the face fi−1 in the open chain and assume the color assigned to the
0
in the expansion E(S, T ) is yellow. Thus the color assigned
corresponding face fi−1
to the face t0i of the expansion corresponding to the edge ti of Γ must be a different
color, say blue. Let gi and di denote the two faces that share the edge ti of Γ. In the
0
and t0i , so gi and di
expansion the corresponding faces gi0 and d0i are adjacent to fi−1
must be in the remaining color class, red. The edge ti also exits the face fi , and so t0i ,
gi0 and d0i are adjacent to fi0 . Thus fi0 must be in the yellow color class. If the chain
makes a sharp turn when entering the next edge ti+1 , then t0i+1 is adjacent to fi0 and
either gi0 or d0i , and thus must be in the blue color class. If there is not a sharp turn,
then by Lemma 2.3, ti+1 exits fi from a vertex on the opposite side of fi , a distance
of three vertices. The faces incident with fi0 alternate in color, so t0i and t0i+1 must be
in the same color class.
2.2
Chains
Let Γ be a fullerene with a chain decomposition (S,T). Let f0 , t1 , f1 , t2 , f2 , ...tk , fk be
an open chain of this decomposition. Each face fi is exited by the edges ti and ti+1
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
di
fi-1
di
di'
fi
gi
fi-1'
ti '
g i'
24
fi '
ti+1'
fi-1
fi
gi
Figure 2.2: Expansion of edges of distance 3 around f with face 3-coloring
in the chain. The initial faces f0 and fk must be pentagons and the remaining fi
can be hexagons or pentagons. We define the augmented chain to be the open chain
together with the faces incident with edges in T over the chain. For each edge ti in
the open chain, we denote the two faces incident with it as di and gi , where di is the
face on the left if we are traversing the chain in increasing order over the indices (see
Figure 2.3). If a chain makes a sharp turn to the left when entering edge ti+1 , then
di = di+1 ; if the chain makes a sharp turn to the right, then gi = gi+1 . We say that
two augmented chains are separated if they do not share any vertices. We say that
an augmented chain is detached if it is separated from every other augmented chain
in this decomposition.
Let Γ be a fullerene with a chain decomposition and an improper face 3-coloring.
We now use the chain decomposition and the improper face 3-coloring to form a new
Kekulé structure for Γ that gives a lower bound for the Clar number and Fries number
over the fullerene.
Choose one of the three color classes and let C be the set of faces in this color
class outside of the augmented chains. We call this set of faces a Clar scheme for Γ.
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
d1
f0
t1
f1
d2
t2
g1
d3
t3
f2
25
f3
g3=g4
g2
t4 d4
f4
g5
f5
t5
d5=d6
t6
g6
d8
d7
f6
t7
g7
f7
t8
g8
f8
Figure 2.3: An augmented chain f0 , t1 , f2 , ..., t8 , f8 between pentagons f0 and f8 . Dark
blue edges represent edges in T .
By Theorem 2.6, all improperly colored faces are contained in the augmented chains,
as are all of the pentagons. Therfore C is an independent set of hexagonal faces,
and every vertex outside of the augmented chains is incident with exactly one face
in C. Given a Clar scheme C, we construct a partial Kekulé structure for Γ in the
following way: For each face in C, choose three alternating bounding edges to be in
the matching. The Clar scheme includes all faces in the chosen color class outside of
the augmented chains, so each vertex outside of the augmented chains is covered by
the matching. We call this matching the Kekulé scheme derived from C.
In the theorem below, we show that given a Clar scheme C with detached chains, it
is possible to extend the Kekulé scheme derived from C over these augmented chains,
possibly adding additional independent benzene faces to the set containing C. Let
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
26
C 0 be a maximal independent set of benzene faces containing C over the extension.
The cardinality of C 0 is a lower bound for the Clar number of the fullerene, and by
Proposition 1.2, |C 0 | =
|V |
6
−
|V ∗ (C 0 )|
6
where V ∗ (C 0 ) is the set of vertices not incident
with a face in C 0 . The contribution to |V ∗ (C 0 )| over a chain depends on the position
of the faces in C relative to the chain. For each case, we find the contribution to
|V ∗ (C 0 )| over the chain. Within an augmented chain f0 , t1 , f1 , ..., tk , fk , we define
non-consecutive sharp turns to be those that do not share a common edge in the
chain.
Theorem 2.8. Let Γ be a fullerene with a chain decomposition (S, T ) into detached
augmented chains. Let C be a Clar scheme and KC the derived Kekulé scheme. Then
it is possible to complete KC to a Kekulé structure K, possibly adding additional
independent benzene faces. Let C 0 denote this enlarged set of independent benzene
faces containing C. Let f0 , t1 , f1 , ..., tk , fk be a detached augmented chain and let di
and gi denote the faces incident with ti in the chain.
1. If the set C is contained in the same color class as the quadrilateral faces {t0i }
in the expansion E(S, T ), then the contribution to |V ∗ (C 0 )| over the chain is 2k.
2. If the set C is contained in the color class that includes the set of faces {di , gi }
and l is the number of non-consecutive sharp turns over the chain, then the
contribution to |V ∗ (C 0 )| is 4k − 6l.
3. If the set C is contained in the color class that includes the faces {fi }, then the
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
27
contribution to |V ∗ (C 0 )| is 6k + 4.
Proof. The Kekulé scheme KC contains three alternating edges bounding each face
in C, so the only vertices not incident with a face in C are over the six augmented
chains. By Lemma 2.7 (1), the improperly colored faces {di , gi : 1 ≤ i ≤ k} over
the augmented chain are all in one color class and the faces fi in the open chain are
all in a second color class over the improper face 3-coloring of Γ. The type of repair
necessary to this chain depends only upon which of the color classes includes the set
C.
Case 1: The set C is contained in the same color class as the quadrilateral faces
{t0i } in the expansion E(S, T ) over the augmented chain. In this case, the only vertices
along the chain not covered by faces in C are the vertices incident with the edges ti .
Adding these k edges to KC extends the Kekulé scheme to cover the vertices of the
chain and contributes 2k vertices to |V ∗ (C 0 )|.
Case 2: The set C is contained in the color class that includes the (improperly
colored) set of faces {di , gi } incident with edges in T along the chain. For each pair
of faces incident with an edge in T , exactly one of the two faces is added to C 0 , the
enlargement of C.
We know from the construction of the coupling in Lemma 2.3 that coupled edges
over a hexagon h exit either from opposite vertices on h (continue without a turn)
or through adjacent vertices (sharp turns). The chain f0 , t1 , f1 , t2 , f2 , ...tk , fk has k
edges; let l be the number of non-consecutive sharp turns. Over segments of the open
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
28
chain with no turns, there are two improperly colored faces for each edge in T and
which of these faces is included in C 0 is inconsequential. The unchosen face has one
edge incident with a face in C 0 and four remaining vertices that are not incident with
a face in C 0 . These vertices are consecutive on the face and can be covered by adding
two Kekulé edges incident with the face. Each non-consecutive sharp turn contains
two edges in T . Therefore, there are k − 2l edges in T that each contribute four
vertices to |V ∗ (C 0 )|.
If the chain makes a sharp turn, there are three improperly colored faces pairwise
incident with two edges in T . Without loss of generality, if the chain makes a sharp
right turn from ti to ti+1 , then the edge ti is incident with the faces di and gi = gi+1
and ti+1 is incident with gi = gi+1 and di+1 . We choose the two faces di and di+1
outside of the turn to be in C 0 . The single interior face gi = gi+1 is chosen not to
be in C 0 . Two edges ti and ti+1 of gi = gi+1 are incident with faces in C 0 , so there
are only two consecutive vertices on this interior face that must be covered by a new
Kekulé edge (see Figure 2.4).
If the chain makes consecutive sharp turns, the interiors of the turns alternate
between faces in C 0 and faces not in C 0 with two uncovered vertices. Each nonconsecutive sharp turn contributes two adjacent vertices to V ∗ (C 0 ) that can be covered
by a single edge in K. On the other hand, in a consecutive pair of sharp turns, the
interior is covered by a face in C 0 and does not contribute any vertices to V ∗ (C 0 ).
The total number of vertices in V ∗ (C 0 ) over the chain is 4(k − 2l) + 2l = 4k − 6l. An
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
29
Figure 2.4: Case 2: The edges in T are dark blue; the faces in C are the yellow faces
not incident with edges in T . The yellow faces with green circles are added to C to
get C 0 . The large red vertices belong to V ∗ (C 0 ). Over this chain, there are k = 16
edges in T and l = 5 non-consecutive sharp turns. Thus the contribution to |V ∗ (C 0 )|
is 4 × 16 − 6 × 5 = 34 .
example is worked out in Figure 2.4.
Case 3: The set C is contained in the color class that includes the set of faces
{fi }. By Lemma 2.2, at least one edge of K exits the pentagon f0 and begins an
open chain that ends with the pentagon fk . First extend KC by adding the edges
t0 , t1 , ...tk to K. Since ti exits fi , fi cannot be a benzene face and thus cannot belong
to C 0 for 1 ≤ i ≤ k; since f0 is a pentagon, f0 cannot belong to C 0 . Hence each vertex
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
30
incident with fi belongs to V ∗ (C 0 ). The faces fi are in the color class that includes
the set C, so no vertex incident with a face fi is covered by a face in C 0 . For each
face fi , the vertices that are not incident with ti or ti−1 are either consecutive or two
adjacent pairs of vertices. In either case, pairs of these vertices can be covered by
edges of K. The vertices in V ∗ (C 0 ) are exactly those incident with the faces fi . There
are k − 1 hexagons and two pentagons in the chain, so the contribution to |V ∗ (C 0 )|
is 6(k − 1) + 10 = 6k + 4.
Let Γ be a fullerene with a given chain decomposition and associated improper
face 3-coloring. A Fries scheme D consists of the faces in one of the three color
classes. In contrast with a Clar scheme, the set D includes the faces of that color
class within the augmented chains. We then construct the Kekulé scheme derived
from D, a partial Kekulé structure consisting of the edges that join two faces from
D. (See Figure 2.5.) Outside of the augmented chains, every vertex is on an edge
joining two faces in D, so all vertices outside of the augmented chains are incident
with exactly one edge in the Kekulé scheme derived from the Fries scheme.
Corollary 2.9. Let Γ be a fullerene with a chain decomposition in which the augmented chains are detached. Let D be a Fries scheme and KD the Kekulé scheme
derived from D. Then there is a Kekulé structure K such that the symmetric difference KD 4K is contained in the augmented chains.
Proof. Consider a fullerene Γ with a chain decomposition in which the augmented
chains are detached. Given a Fries scheme D and the Kekulé scheme KD derived
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
31
(a) Here the Fries scheme, D, consists of the faces in the blue color class.
(b) Here the Fries scheme, D, consists of the faces in the pink color class.
Figure 2.5: The Kekulé scheme derived from D consists of edges joining two faces in
D. Dark blue edges represent Kekulé edges, red edges represent edges in T . In each
case, all vertices outside of the augmented chain are covered by exactly one edge in
the Kekulé scheme.
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
32
from D, choose a second color class to contain a Clar scheme C, the set of hexagons
within the second color class and outside of the augmented chains. Construct a Kekulé
scheme KC derived from C consisting of three alternating edges on each face of C,
choosing the edges that join two faces in D. Outside of the augmented chains, this
KC is identical to KD : each of the edges in KC connects two faces in the color class
containing D. Since every face in the second color class is included in C outside of
the augmented chains, each of the vertices outside of the chains are covered by edges
from KC . By Theorem 2.8, we can extend KC to a Kekulé structure K while only
changing edges inside of the augmented chains. Thus K and KD differ only within
the augmented chains.
In summary, Corollary 2.9 shows that we can choose a set of faces D to be the
Fries scheme, we can create a Kekulé scheme KC based on C with the Fries scheme
in mind: for each face in C, we have two choices for the set of alternating bounding
edges to include in KC . We always choose faces that connect two faces in D. Since
outside of the chains, KC is equivalent to the Fries scheme we would derive directly
from D, extending KC to a Kekulé structure using the method of Theorem 2.8 forms
a Kekulé structure that differs from a Kekulé scheme derived directly from D only
within the augmented chains. We can then calculate a lower bound for the Fries
number from this Kekulé structure. This is not necessarily the best Kekulé structure
based on a Fries scheme; the corollary is showing only that one can be completed by
only repairing edges within the augmented chains.
CHAPTER 2. CHAINS AND IMPROPER FACE 3-COLORINGS
33
Theorem 2.8 and Corollary 2.9 show that when a fullerene admits a chain decomposition with detached chains, we can construct Kekulé structures where the only
contributions to the Clar and Fries deficits come from the augmented chains.
34
Chapter 3
Clar Structures
3.1
Clar Structures
Chapter 2 describes a method for calculating lower bounds for the Clar number of
a fullerene through a chain decomposition. In this chapter we construct a Kekulé
structure that achieves the Clar number for a fullerene and see that this structure
comes from a chain decomposition. We begin in a new general setting to describe
this construction.
Define a vertex covering (C, A) of a 3-regular plane graph Ω = (V, E, F ) to be a
set of faces C and edges A such that all vertices of Ω are incident with exactly one
element of C ∪ A.
Lemma 3.1. Let Ω be a plane graph with a vertex covering (C, A). On every face of
even degree, there is an even number (possibly zero) of edges in A that exit the face.
CHAPTER 3. CLAR STRUCTURES
35
On every face of odd degree that does not belong to C, there is an odd number of edges
in A that exit the face.
Proof. Let Ω be a plane graph with a vertex covering (C, A) and let f be a face of Ω
that is not in C. Note that every edge in A that lies on f covers two adjacent vertices
on f , as does every face in C adjacent to f . The proof proceeds exactly as in Lemma
2.2 of Chapter 2.
The vertex covering (C, A) is a face-only vertex covering if the set A is empty.
Lemma 3.2. If C is a face-only vertex covering of a plane graph Ω, then C must
contain all of the faces of Ω with odd degree.
Proof. We see in Lemma 3.1 that if p is a face of odd degree in Ω and p is not in a
vertex covering (C, A), then at least one edge from A exits p, so A 6= ∅.
Fowler and Pisanski referred to a face-only vertex covering in a fullerene Γ as a
perfect Clar structure [5]. We define a Clar structure (C, A) of a fullerene Γ to be a
vertex covering where C consists only of hexagonal faces and at most two edges in A
lie on any face of Γ. Given a Clar structure (C, A), choose three alternating edges on
each face in C. Together with the edges in A, these edges form a Kekulé structure K.
We say that K is a Kekulé structure associated with the Clar structure (C, A). Note
that the faces in C form a maximal independent set of benzene faces. Conversely,
given a fullerene Γ with a Kekulé structure K, we can form a Clar structure (C, A)
associated with K: take C to be a maximal independent set of benzene faces and A
CHAPTER 3. CLAR STRUCTURES
36
Figure 3.1: A benzene chain. Edges in A exit yellow faces from adjacent vertices to
form a short closed chain.
to be the remaining edges in K. The Clar number of a fullerene is given by a Clar
structure (C, A) with a maximum number of faces in C.
Lemma 3.3. Let Γ be a fullerene with |V | vertices and a Clar structure (C, A). Then
|C| =
|V | |A|
−
.
6
3
Proof. We see in Proposition 1.2 that for an independent set C of benzene faces in a
Kekulé structure K, |C| =
|V |
6
− |V
∗ (C)|
6
where V ∗ (C) is the set of vertices not incident
with a face in C. In a Clar structure (C, A), 2|A| = |V ∗ (C)|.
We want to maximize the number of independent benzene faces, and we know
that |C| =
|V |
6
− |A|
, so our goal is to find a Clar structure (C, A) that minimizes |A|.
3
Given a fullerene Γ with Clar structure (C, A), define a Clar chain to be an open or
closed chain using only edges from A. Let K be a Kekulé structure associated with
(C, A). For any face in C, we call the three edges in K bounding this face together
with the three faces exited by these edges a benzene chain (see Figure 3.1).
CHAPTER 3. CLAR STRUCTURES
37
Lemma 3.4. Let (C, A) be a Clar structure for a fullerene Γ. Let K be a Kekulé
structure associated with (C, A). Then there is a non-crossing coupling for K such
that:
1. The chains are either Clar chains or benzene chains corresponding to the faces
in C.
2. The coupling of edges in A can be chosen so that all edges coupled around pentagons exit from adjacent vertices on the pentagon.
3. There are six open Clar chains connecting pairs of pentagons.
Proof. Let (C, A) be a Clar structure and let K be a Kekulé structure associated with
(C, A). We want to show that there is a non-crossing coupling of the edges in K that
has the properties above. We choose this coupling in accordance with Lemma 2.3 to
ensure that the edges in A do not cross one another. In this lemma, the first step to
couple the edges around a hexagon is to choose a pair of edges exiting the face from
adjacent vertices. Around a pentagon, the first step is to choose the initial exit edge,
and the second is to couple a pair of edges exiting from adjacent vertices.
Consider the three Kekulé edges bounding a face in C. These are not edges in A,
and therefore none can be the initial edge of a pentagon. Any two of these edges exit
a common face from adjacent vertices of that face. Thus for each face in C we can
couple the edges together to form a benzene chain. Only the edges in A remain to be
coupled. For each hexagon h of Γ, an even number of edges from A exit h by Lemma
CHAPTER 3. CLAR STRUCTURES
38
Figure 3.2: For Clar chains, edges in A coupled over pentagons must exit from adjacent vertices. The indicated couplings are avoided.
3.1. We can couple these edges around h in accordance with Lemma 2.3.
Consider a pentagon p of Γ. If a single edge from A exits p, this is the initial
edge and is not coupled over p. If more than one edge in A exits p, we know that
the edges from A exit from adjacent vertices on p. Couple the edges in the following
way: Choose edges exiting from adjacent vertices on p to be coupled. If a single
edge remains, it is the initial edge of the pentagon. If three remain, couple two edges
exiting from adjacent vertices and let the remaining edge be the initial edge of the
pentagon. From chapter 2, we see that there must be six open Clar chains connecting
pairs of pentagons.
This method is more restrictive than that of Lemma 2.3, because coupled edges
over pentagons only exit from adjacent vertices, ensuring (2) (see Figure 3.2).
CHAPTER 3. CLAR STRUCTURES
39
Let Γ be a fullerene with a Clar structure (C, A) with an associated Kekulé structure K and a coupling for K as in Lemma 3.4. We define the expansion E(C, A) to
be the graph obtained by the following operation: widen the edges in A into quadrilateral faces as described in Section 2.1. Each vertex incident with an edge in A
becomes an edge, and each edge in A splits lengthwise into two edges. Note that
the expansion in Chapter 2 only expanded edges contained in open chains; here there
may be additional edges in A contained in closed Clar chains. The advantage is that
the set C of independent hexagons is preserved; the faces of C and the quadrilateral
faces in E(C, A) corresponding to edges from A form a color class in E(C, A). Thus
in the associated improper face 3-coloring for Γ, no faces in the set C are improperly
colored.
Lemma 3.5. Let Γ be a fullerene with a Clar structure (C, A). The expansion E(C, A)
is face 3-colorable. All faces of E(C, A) corresponding to faces in C and edges in A in
Γ are in one color class of E(C, A). Furthermore, Γ has an associated improper face
3-coloring for which the only improperly colored faces are those that share edges in A.
For an open or closed Clar chain f0 , a1 , f1 , a2 , f2 , ...ak , fk in this improper 3-coloring,
the faces fi are all in one of the remaining two color classes, and the faces incident
with edges in A over the chain are all in the third color class.
Proof. Let f be a face of degree d in the fullerene Γ. If j edges from A exit f , then
the corresponding face in E(C, A) has degree d + j. By Lemma 3.1, pentagons in
Γ are exited by an odd number of edges from A, hexagons by an even number of
CHAPTER 3. CLAR STRUCTURES
40
edges from A. Thus every face in E(C, A) is of even degree, and E(C, A) is face 3colorable by Theorem 2.1. Since (C, A) is a vertex covering of Γ, the faces of E(C, A)
corresponding to the faces in C and the edges in A comprise one color class. Suppose
that these faces are blue. For any open or closed chain f0 , a1 , f1 , a2 , f2 , ...ak , fk (where
fk = f0 if the chain is closed), the argument given in Lemma 2.7 shows that the faces
fi are all in a second color class (red or yellow). Each face sharing an edge ai over
the chain is in the third color class (yellow or red).
Let f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 be a closed Clar chain. The fullerene is now
partitioned into three parts: the chain f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 , and two regions.
We say that the region containing the least number of pentagons is the interior of
the chain and the region on the other side of the chain is the exterior. If both sides
contain the same number of pentagons, we arbitrarily choose one region to be the
interior. Define a chain circuit C around the interior to be an elementary circuit
containing the edges ai connected by paths on the boundaries of the fi . There are
many possible circuits for the closed chain f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 , depending
on whether we connect each edge ai to ai+1 by a path of vertices on fi incident with
the interior or the exterior of the Clar chain for each i. For a chain circuit C around a
closed Clar chain f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 , we define the interior of C to be the
interior of f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 together with the faces fi that do not share
an edge of C with the interior of the chain f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 .
Lemma 3.6. Let Γ be a fullerene with a Clar structure (C, A) and an associated cou-
CHAPTER 3. CLAR STRUCTURES
41
pling. If |C| is the Clar number for Γ, then every closed Clar chain f0 , a1 , f1 , a2 , f2 , ...ak , fk =
f0 has a pentagon in its interior.
Proof. Consider a closed Clar chain with no pentagons in its interior. There may
additionally be nested closed chains in the interior; let f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0
be the innermost closed Clar chain. For i = 1, 2, ..., k, let gi and di be the faces
incident with ai on the interior and exterior of the chain, repectively. Give Γ the
improper face 3-coloring associated with (C, A), and assume that C is contained in
the blue color class. By Lemma 3.5, all faces fi belong to a second color class, say
yellow, and all gi and di are in the remaining color class (red). By Lemma 3.4 Clar
chains are non-crossing. Consequently, there are no edges from A in the interior
of this innermost Clar chain. Thus by Lemma 3.5, all faces are in the interior of
f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 properly 3-colored.
Consider two faces gi and di that share an edge ai of the chain. The faces gi and
di are red, and each is adjacent to the yellow faces fi−1 and fi . The interior face gi
conforms with the proper face 3-coloring of the interior of the chain. If the interior
coloring is extended to the exterior face di , then di would be a blue face. We see
that this is the case for all of the improperly colored faces dj , gj incident with edges
aj over the chain for 1 ≤ j ≤ k. The faces {dj } together with the set of blue faces
in the interior of the chain form an independent set. If we interchange the blue and
red color classes within the interior of the chain, then the set of faces in the interior
together with the faces {dj } are properly face 3-colored. All vertices on the chain
CHAPTER 3. CLAR STRUCTURES
42
circuit and its interior are then incident with exactly one blue face. Now let C 00 be
the collection of all blue faces and let A00 = A \ {a1 , a2 , ..., ak }. Now (C 00 , A00 ) is a Clar
structure, and |C 00 | > |C| by Lemma 3.3. Therefore, if (C, A) is a Clar structure that
attains the Clar number, then a closed chain containing only hexagons in its interior
cannot exist.
Lemma 3.7. A chain circuit C is of even length if and only if its interior contains
an even number of pentagons.
Proof. Suppose that there are h hexagonal faces and p pentagonal faces in the interior
of C. If an edge of one of these faces is on the circuit, then the edge is incident with
only one of these faces. If the edge is in the interior, it is incident with two of these
faces. Let I be the number of edges in the interior. Then the number of edges on the
circuit is 6h + 5p − 2I = 2(3h − I) + 5p. So the number of edges on the circuit is even
exactly when p is even.
Lemma 3.8. Let f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 be a closed Clar chain such that each
of the fi ’s are hexagons. Then all of its chain circuits have even length.
Proof. By the construction of the coupling for the Clar structure (C, A), the paired
edges exiting a hexagon fi exit from adjacent vertices of fi (a sharp turn) or from
vertices on opposite sides of f . The distance traveled by the chain circuit from the
edge ai to ai+1 is odd (distance 1, 5, or 3). As the chain circuit goes around a hexagon
fi from ai to ai+1 , it passes through an even number of vertices not included in the
coupled edges a1 , a2 , ..., ak . Thus the chain circuit is of even length.
CHAPTER 3. CLAR STRUCTURES
43
Lemma 3.9. For any closed Clar chain, there exists a chain circuit that includes an
even number of pentagons on its interior.
Proof. By Lemmas 3.7 and 3.8, if each of the faces fi in the closed Clar chain
f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 is a hexagon, then there is an even number of pentagons on the interior regardless of how the chain circuit is chosen. Suppose that
there is an odd number of pentagons in the interior of the chain. By Lemma 3.7,
some face fj must be a pentagon. We can choose the chain circuit to traverse the
outside of this pentagon, and the inside of every other face fi on the chain. We have
chosen a chain circuit for f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 that contains an even number
of pentagons in its interior.
In Chapter 2, to define an improper face 3-coloring, the closed chains are ignored,
and we define a face 3-coloring for the graph E(Γ) obtained by expanding only the
edges in open chains joining pentagons and then choosing a color class for the Clar
faces. If we ignore the closed chains here, it may be the case that none of the resulting
color classes contains C. Hence we cannot ignore closed chains in this chapter.
We have shown that in a coupling of a Clar structure (C, A), the Clar chains are
a special case of the chains in Chapter 2, and thus the lemmas already proven about
chains hold. Define a straight chain segment to be an alternating chain f0 , a1 , f1 , a2 , ...ak , fk
of edges in A and hexagons fi such that the edges ai and ai+1 exit from opposite vertices of fi for each i. A straight chain segment with k edges in A connects a pair
of faces with Coxeter coordinates (k, k). Any edge in A is part of a straight chain
CHAPTER 3. CLAR STRUCTURES
44
Figure 3.3: A straight chain segment with six edges in A joining a pair of faces with
Coxeter coordinates (6, 6). Dark blue edges represent edges in A.
segment of length at least one. Thus any Clar chain between two faces is a sequence
of straight chain segments. We can visualize a straight chain segment with k edges
as the diagonal of a parallelogram with edges of length k through faces and with the
straight chain along the diagonal of the parallelogram (see Figure 3.3).
Lemma 3.10. Every open Clar chain in a fullerene Γ with Clar structure (C, A) is
a sequence of straight chain segments with only sharp turns.
Proof. By Lemma 3.4, the coupling is chosen so that edges in A coupled over a
pentagon exit from adjacent vertices. Thus Clar chains always make sharp turns as
they pass through pentagons. Because the coupling is non-crossing, coupled edges
through a hexagon exit from adjacent vertices (a sharp turn) or vertices on opposite
sides of the hexagon (a continuation of a straight chain segment).
CHAPTER 3. CLAR STRUCTURES
45
We say that a Clar structure (C, A) has a coupling with non-interfering Clar
chains if for every pair of pentagons p1 and p2 joined by a Clar chain there is no other
pentagon that has a vertex in common with any shortest chain joining p1 and p2 in
any coupling of (C, A). For the rest of this paper, we consider pairs of pentagons
that can be joined by non-interfering Clar chains. When more than two pentagons
interact, chains can become quite complicated; a small example is given in Figure
3.4(a) containing pentagons p1 , p2 , p3 , and p4 . In this example, nearby pentagons p1
and p2 are in different color classes and cannot be joined directly with Clar chains
by Lemma 3.5. We must instead take a Clar chain between p1 and p3 that twists
around the pentagon p4 . If we consider a similar patch that does not contain the
pentagons p2 and p4 (see Figure 3.4(b)), then the pentagons p1 and p3 are in different
color classes and cannot be connected by a Clar chain. The ability to connect a pair
of pentagons by a Clar chain can depend upon the positions of other pentagons when
we consider the more general case.
Restricting our attention to non-interfering Clar chains still permits us to solve
two major open problems: to classify fullerenes that attain the theoretical maximum
for the Clar number, as well as to find classes of fullerenes for which the Clar and
Fries number cannot be attained with the same Kekuklé structure. We are also able
to compute the Clar number for several classes of fullerenes. In future research, we
aim to develop an understanding of Clar chains more generally, but for now, we focus
on non-interfering chains.
CHAPTER 3. CLAR STRUCTURES
46
p1
p1
p2
p4
p3
p3
(a) A Clar chain connecting pentagons p1 and p3 , another (b) When we remove the other two penconnecting pentagons p2 and p4 .
tagons, p1 and p3 cannot be connected by
a Clar chain. White faces represent faces
that cannot be colored in our improper
face 3-coloring.
Figure 3.4: Chains can become complicated when more than two pentagons interact.
Lemma 3.11. Assume two pentagons are joined by a non-interfering Clar chain.
Then any shortest Clar chain joining them is composed of alternating right and left
turns.
Proof. If a Clar chain takes two consecutive right turns, we have turned 120◦ and
are traveling toward the first straight chain segment. Thus a face reached by two
consecutive right turns could be reached by two shorter segments.
Suppose the chain connecting pentagons p1 and p2 consists of a straight chain
segment with k edges in A, then a sharp left turn followed by a straight chain segment
with l edges in A. Begin with the pentagon p1 . The first straight chain segment goes
to the face with Coxeter coordinates (k, k). The side of the second parallelogram goes
backwards 2l faces along the side of the first parallelogram (in the first coordinate)
CHAPTER 3. CLAR STRUCTURES
47
and l faces in the positive direction in the second coordinate. If 2l ≤ k, the second
parallelogram ends at p2 with coordinates (k − 2l, k + l). (See Figure 3.5(a).) If we
instead have a sharp right turn and 2l ≤ k, then the coordinates are reversed, and
the segment has Coxeter coordinates (k + l, k − 2l).
If 2l > k ≥ l, then going backwards 2l faces in the first coordinate after a sharp
left turn takes us to k −2l < 0. Since this is negative, we re-orient the segment so that
it is in the positive direction, and we have 2l − k as our second Coxeter coordinate.
In the case of a sharp left turn, going backwards 2l faces takes us past the point (k);
it takes us to the coordinate (2k − 2l, 2l − k). Going forwards l faces in the now-first
coordinate takes us to (2k − l, 2l − k). (See Figure 3.5(b).) For a sharp right turn,
the Coxeter coordinates of the segment are (2l − k, 2k − l).
Lemma 3.12. Suppose two faces of a fullerene Γ can be connected by a Clar chain
and the orientations of the parallelograms are given. If the segments alternate between
right and left turns, the sum of the edges in A over any such Clar Chain is the same
regardless of the number of turns.
Proof. If a chain alternates between right and left turns, the jth parallelogram is in the
same orientation as the first parallelogram for j odd, and in the same orientation as the
second for j even since all turns are at 60◦ angles. Any parallelograms with the same
orientation are adding edges to A in the same direction. Thus one parallelogram with
k edges in |A| reaches the same face as r parallelograms with k1 , k2 , ...kr parallelograms
in the same orientation such that k1 + k2 + ... + kr = k. So a parallelogram with k
CHAPTER 3. CLAR STRUCTURES
48
(5-4, 5+2)
p2
(5, 5)
(10-4, 8-5)
(5, 5)
p4
p1 (0, 0)
p3 (0, 0)
(a) A straight chain segment of length k =
(b) A straight chain segment of length k = 5 fol-
5 followed by a straight chain segment of
lowed by a straight chain segment of length l = 4.
length l = 2. The Coxeter coordinates be-
The Coxeter coordinates between p3 and p4 are
tween p1 and p2 are (1, 7).
(6, 3).
Figure 3.5: Pairs of pentagons connected by two straight chain segments. The pentagonal faces are yellow.
edges in A followed by a parallelogram with l edges in A reaches the same point
as any number of parallelograms with alternating turns such that diagonals of the
odd-numbered parallelograms add up to k and the diagonals of the even-numbered
parallelograms add up to l.
CHAPTER 3. CLAR STRUCTURES
3.2
49
Calculating the Number of Edges in A over a
Clar Chain
To find the Clar number for classes of fullerenes, we would like to calculate the number
of edges in A over Clar structures (C, A) for these fullerenes, since |C| =
|V |
6
−
|A|
.
3
In this section, we calculate |A| over non-interfering Clar chains. We show that if
two pentagons can be connected by a non-interfering Clar chain, then there is a Clar
chain between them consisting of at most two straight chain segments. By Lemmas
3.11 and 3.12, any chain that alternates between right and left turns contributes
a minimum number of edges to A. Thus a chain with a single turn contributes a
minimum number of edges to A.
To describe Clar chains in the non-interfering case, consider two pentagons p1 and
p2 that are joined by a Clar chain. Start with the hexagonal tessellation and at each
pentagon, cut out a 60◦ wedge and identify the rays bounding the wedge. Assign the
faces the improper face 3-coloring given by the chain. The chain and the two wedges
split our region and the face 3-colorings above and below the split must match when
the wedges are collapsed (see Figure 3.6). For any Clar structure (C, A), the edges
in A in the Clar chain between p1 and p2 together with the faces in C must form a
vertex covering over this patch.
Let p1 and p2 be pentagons connected by a Clar chain. Suppose that the Coxeter
coordinates of the segment between p1 and p2 are (m, n). We can assume without loss
CHAPTER 3. CLAR STRUCTURES
50
P1
P2
Figure 3.6: The Coxeter coordinates of the segment between pentagons p1 and p2 are
(8, 2). If the yellow faces contain the set C, this chain is of Type 1. If the red faces
contain the set C, the chain is of Type 2. If the blue faces contain the set C, this
chain is of Type 3.
of generality that m ≥ n. Begin with p1 at the origin and consider a parallelogram
with sides parallel to the directions of the Coxeter coordinates (see Figure 3.6). The
chain is of Type 1 if all four of the vertices of p1 outside of the parallelogram are
covered by a face in C. The chain is of Type 2 if exactly two of the vertices outside
of the parallelogram are covered by a face in C. The chain is of Type 3 if none of the
vertices outside of the parallelogram is covered by a face in C.
CHAPTER 3. CLAR STRUCTURES
51
Lemma 3.13. Let p1 and p2 be two pentagons joined by a chain of Type 1 and let
(m, n) (or (m)) be Coxeter coordinates of the segment between them, where m ≥ n.
Then this chain contributes m edges to A and m ≡ n (mod 3).
Proof. Suppose the first straight segment has a edges in A and the second has b
edges in A. The Coxeter Coordinates between p1 and the face at the sharp turn
are (a, a). The next straight segment has Coxeter Coordinates (b, b), and travels
backwards b faces in the direction of the second coordinate, arriving at a face with
coordinates (a, a − b). Our final step goes b steps in the first coordinate and another
b steps backwards in the second coordinate. The Coxeter coordinates of the segment
between p1 and p2 are then (a + b, a − 2b), which we know is equal to (m, n). Since
a + b and a − 2b are congruent modulo 3, m ≡ n (mod 3). Solving for a and b we find
that a =
n+2m
3
and b =
m−n
.
3
The number of edges in the two straight chain segments
is a + b = m. This argument works in the special case where the segment between p1
and p2 has Coxeter coordinate (m) since (m) is equivalent to the Coxeter coordinates
(m, 0).
A completed chain of Type 1 is shown in Figure 3.7(a).
Lemma 3.14. Let p1 and p2 be two pentagons joined by a chain of Type 2 and let
(m, n) be Coxeter coordinates of the segment between p1 and p2 , where m ≥ n. This
chain contributes m + n edges to A and m ≡ n (mod 3).
Proof. By definition of Type 2, exactly two of the vertices of p1 are covered by a
face in C. Let v1 , v2 , v3 , v4 , v5 be the vertices bounding p1 in clockwise order, with
CHAPTER 3. CLAR STRUCTURES
52
v2
P1
P1
v1
P2
P2
(a) A chain of Type 1 connecting pentagons p1 and
(b) A chain of Type 2 connecting pentagons p1 and
p2 . The blue faces contain the set C.
p2 . The red faces contain the set C. Vertices v1
and v2 are the vertices of p1 incident with a face
in C outside the parallelogram.
Figure 3.7: Chains of Type 1 and 2. Thick blue edges are edges in A.
v1 and v2 incident with a face in C. A chain of Type 1 is not possible: the first
edge would exit from the vertex v4 , and the vertices v3 and v5 would not be covered
by an element of C ∪ A. The first edge in A must exit p1 from v3 or v5 , and we
cover the remaining two adjacent vertices with another face in C. Thus there are two
choices for the direction of the first parallelogram; by symmetry, the direction chosen
is inconsequential.
Choosing to exit v5 , we use two straight chain segments to reach p2 with the first
CHAPTER 3. CLAR STRUCTURES
53
h1 p1
Figure 3.8: A chain of Type 3 begins with a pentagon p1 and must also start a closed
Clar chain through the hexagon h1 . Here the blue faces contain the set C, the thick
edges represent edges in A.
containing a edges in A and the second containing b edges in A, where a ≥ b. (See
Figure 3.7(b).) If we begin with the coordinates (0, 0), the coordinates at the end of
the first parallelogram are (a, a), and we are at distance (b, b) from p2 . When traveling
along the next parallelogram, we go backwards 2b faces along the first parallelogram
(in the direction of the second coordinate) and b faces in the direction of the first
coordinate. The difference from Type 1 is that here, 2b > a. We end with the
coordinates (2a − b, 2b − a)=(m, n). Solving for a and b gives a =
2m+n
3
and b =
2n+m
3
and hence a + b = n + m. Since m − n = 3a − 3b, m ≡ n (mod 3).
A completed chain of Type 2 is shown in Figure 3.7(b).
Lemma 3.15. Let p1 and p2 be two pentagons joined by a chain of Type 3 and let
(m, n) be Coxeter coordinates of the segment between them, where m ≥ n. Then this
CHAPTER 3. CLAR STRUCTURES
54
chain contributes 3m + 2 edges to A and m ≡ n (mod 3).
Proof. By definition of Type 3, none of the vertices of p1 is covered by a face in C.
This means that p1 and p2 are in the independent set of faces containing C. Thus all
the vertices of the pentagon must be covered by edges in A. If exactly one edge in A
exits the pentagon, then the other four vertices of the pentagon are covered by two
edges in A lying on the pentagon. These two edges exit a hexagon h1 and are coupled
over h1 , then continue a Clar chain with two adjacent hexagons each adjacent to p1
(see Figure 3.8). The Clar chains beginning with h1 end with a hexagon h2 adjacent
to p2 . The union of the two chains connecting h1 and h2 form a closed chain around
the interior chain connecting p1 and p2 . Because p1 is not adjacent to any face in C,
we can choose the direction of the Clar paths to be in the direction of the first Coxeter
coordinate. The coordinates between the h1 and h2 are also (m, n). The interior chain
connecting the pentagons p1 and p2 is a chain of Type 1; the interior consists of two
straight chain segments of lengths a and b, where a + b = m. By Lemma 3.13, m ≡ n
(mod 3), and the interior Clar chain joining p1 and p2 contributes m edges to |A|.
The Clar chains between h1 and h2 are parallel to the open chain. One chain is of the
same length, and the other travels around the parallelogram, so its length is a + b + 2.
The total number of edges in A is 3m + 2. This proof works in the special case where
the segment between p1 and p2 has Coxeter coordinate (m) since (m) is equivalent to
the Coxeter coordinates (m, 0).
To show that 3m + 2 edges are needed, note that the shortest Clar chain between
CHAPTER 3. CLAR STRUCTURES
55
P1
P2
Figure 3.9: A chain of Type 3 connects pentagons p1 and p2 . Thick blue edges
represent edges in A.
pentagons with coordinates (m, n) is of length m. This chain contains m−1 hexagons
and two pentagons. We know that none of the vertices on these faces is covered by
a face in C, so each of these vertices must be covered by an edge from A. There are
6(m − 1) vertices from the m − 1 hexagons and ten vertices from the two pentagons.
These 6m + 4 vertices need 3m + 2 edges in A to cover them.
Lemma 3.16. Let Γ be a fullerene with a Clar structure (C, A) and a coupling. If
|C| is the Clar number for Γ, then any closed Clar chain with exactly two pentagons
p1 and p2 in its interior together with the open Clar chain connecting the pentagons
CHAPTER 3. CLAR STRUCTURES
56
contributes the same number of edges to A as a Clar chain of Type 3 between p1 and
p2 .
Proof. Let f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 be a closed Clar chain with exactly two pentagons p1 and p2 in its interior. By Lemma 3.4, there are six open Clar chains
connecting pairs of pentagons, and the Clar chains are non-crossing. Thus there
must be some open Clar chain p1 , e1 , h1 , e2 , h2 , ...el , p2 joining p1 and p2 that is contained completely inside the closed Clar chain f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 . Give Γ
an improper face 3-coloring associated with the Clar structure (C, A). By Lemma
3.5, we can assume without loss of generality that the faces of C are blue in this face
3-coloring, and that the faces f0 , f1 , f2 , ...fk = f0 of the closed chain are red.
Consider the case in which the faces p1 , h1 , h2 , ...p2 of the open chain are also in the
red color class. We can remove the closed chain by interchanging the colors blue and
yellow in the interior. After this change, the pentagons p1 and p2 are still red. Since
they are not in the color class containing C, the chain between the pentagons is of
Type 1 or Type 2. We know from Lemmas 3.13 and 3.14 that for a pair of pentagons
with Coxeter coordinates (m, n), m ≥ n, a chain of Type 1 contributes m edges to A
and a chain of Type 2 contributes m + n edges. A closed chain surrounding them has
at least 2m edges in A, and there remain edges from A in the open chain. Therefore,
deleting the closed chain and interchanging the interior face colors decreases A. Such
a closed chain does not exist in a Clar structure that attains the Clar number.
Consider the case in which the faces of the open chain p1 , h1 , h2 , ...p2 are in the
CHAPTER 3. CLAR STRUCTURES
57
yellow color class. Removing the closed chain interchanges the colors of the blue and
yellow faces, so p1 and p2 would be in the blue color class. Since the pentagons cannot
be in the set C, we no longer have a vertex covering (C, A). This results in a chain
of Type 3: there are 3m + 2 edges in A connecting the pentagons, and all vertices of
the pentagons are covered by edges in A. The open chain p1 , e1 , h1 , e2 , h2 , ...el , p2 is in
the interior of f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 . Thus if the coordinates between p1 and
p2 are (m, n) with m ≥ n, then an open chain containing these pentagons must have
length at least 2(m + 1). Therefore, the open chain p1 , e1 , h1 , e2 , h2 , ...el , p2 together
with the closed chain f0 , a1 , f1 , a2 , f2 , ...ak , fk = f0 contributes at least 3m + 2 edges
to A.
A chain of Type 3 occurs when the independent set containing C also contains the
paired pentagonal faces of an open chain. To change the color class of the pentagons,
we use a closed chain and then connect the pentagons with a chain of Type 1.
Let Γ be a fullerene that allows a coupling of a Clar structure with non-interfering
Clar chains. Let (mi , ni ) be the Coxeter coordinates of the segments between each
pair of pentagons connected by a Clar chain in this coupling for 1 ≤ i ≤ 6. For an
arbitrary Clar structure of Γ, there are three choices for the set of faces containing C
outside of the six clear fields containing the pairs of pentagons. These three choices
give three distinct Clar structures. Given one of these Clar structures, the number of
edges contributed to |A| by the pair with Coxeter coordinates (mi , ni ) is max{mi , ni }
if the chain is of Type 1, mi + ni if the chain is of Type 2, and 3 max{mi , ni } + 2 if the
CHAPTER 3. CLAR STRUCTURES
58
chain is of Type 3. For each of the three choices, we determine the total number of
edges in |A| and let M be the minimum of these three sums. We define such a pairing
of pentagons to be widely separated over the fullerene if for any two pentagons that
are not paired together, one of the Coxeter coordinates of the segments joining them
is at least
M
2
− 2.
Lemma 3.17. Let Γ be a fullerene over which the pairs of pentagons are widely
separated. If (C, A) is a Clar structure that attains the Clar number for Γ, then
(C, A) includes no closed chain with more than two pentagons in its interior.
Proof. Suppose (C, A) is a Clar structure that attains the Clar number for Γ, and that
(C, A) includes a closed Clar chain f0 , a1 , f1 , ..., ak , fk = f0 with three or more pentagons in its interior. By Lemma 3.9, there exists a chain circuit C for f0 , a1 , f1 , ..., ak , fk =
f0 that includes an even number of pentagons in its interior. Thus at least four
pentagons are in the interior of C. By definition of widely separated, the interior
of C contains at least two pairs of pentagons for which one of the Coxeter coordinates of the segments joining them is at least
M
2
− 2. Therefore the length of
f0 , a1 , f1 , ..., ak , fk = f0 is at least 2( M2 − 2) + 2 = M − 2. There are additional edges
in A for each of the six chains between paired pentagons, so the total contribution to
|A| is greater than M . Thus, (C, A) does not attain the Clar number for Γ.
Theorem 3.18. Let Γ be a fullerene over which the pairs of pentagons are widely
separated. Let M be the minimum sum of the edges in A over the three possible choices
for the face set containing C. Then
|V |
6
−
M
3
is the Clar number for Γ.
CHAPTER 3. CLAR STRUCTURES
59
Proof. By Lemma 3.4, any Clar structure (C, A) over Γ contains six Clar chains
connecting pairs of pentagons. Let M denote the minimum number of edges in A
over the three possible choices for the set C outside of the six clear fields. If a different
pairing is chosen for any pentagon, then there are at least two new pairs of pentagons.
For each new pair, one of the Coxeter coordinates is at least
M
2
− 2 since the original
chains were widely separated. Chains connecting the other four pairs each contribute
at least one edge to A, giving a total of at least M edges in A. By Lemma 3.17, any
closed Clar chain containing more than one pair of pentagons increases the size of A.
Thus any other Clar structure contains at least M edges in A. The conclusion follows
by Lemma 3.3.
60
Chapter 4
Fullerenes with Complete Clar
Structure
In this chapter, we describe a class of fullerenes for which the Clar number is maximal
with respect to the number of vertices in Γ. Let Γ be a fullerene with a Clar structure
(C, A). By definition the cardinality of C is a lower bound for the Clar number of a
fullerene, and we proved in Lemma 3.3 that |C| =
|V | |A|
−
. To describe a class of
6
3
fullerenes for which |C| is maximal with respect to the number of vertices, we find
fullerenes for which |A| is minimal. By Lemma 3.1 at least one edge from A exits
each of the twelve pentagons. There are additional edges in A if one of these edges
exits a hexagon. Therefore |A| ≥ 6, and equality holds exactly when all edges from
A connect pairs of pentagons. We say that a Clar structure is complete if |A| = 6, in
which case |C| =
|V |
6
− 2. We can summarize the above in the following lemma:
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
61
Figure 4.1: Two pentagons paired by a single edge in A
Lemma 4.1. If a fullerene Γ has a complete Clar structure (C, A), then there exists
a pairing of the twelve pentagonal faces such that each pair is connected by a single
edge. The only edges in A are the six edges between pairs of pentagonal faces.
Ye and Zhang called fullerenes with complete Clar structure extremal fullerenes
[12]. They went on to construct the 18 extremal fullerenes on 60 vertices.
4.1
The Leapfrog Construction
The leapfrog construction was described in Chapter 1. By Proposition 1.4, a fullerene
Γ = (V, E, F ) attains the Fries number
|V |
3
if and only if Γ has a face-only vertex
covering. The Kekulé structure that attains the Fries number
|V |
3
is constructed by
taking the faces in the face-only vertex covering to be the void faces. By Lemma
3.1, all of the pentagons of Γ must be in this face-only vertex covering. We say
that a fullerene with these properties has a complete Fries structure. We now use
the leapfrog construction in a different way to construct the fullerenes that admit a
complete Clar structure.
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
62
Let Ω` be the leapfrog of the plane graph Ω = (V, E, F ). By Lemma 1.3, the faces
F ` of Ω` corresponding to the faces F of Ω form a face-only vertex covering of Ω` .
The set of edges that do not bound a face in F ` forms a perfect matching of Ω` , that
is, a Kekulé structure if Ω` is a fullerene. Extending this notation, we write V ` for
the set of faces in Ω` corresponding to the vertex set V of Ω and U ` to denote a set
of faces of Ω` corresponding to a subset U of V .
Lemma 4.2. Given a 3-regular plane graph Ψ with a face-only vertex covering Q,
there is a unique plane graph Ω = (V, E, F ) such that Ω` =Ψ and F ` = Q.
Proof. Consider the planar dual Ψ∗ of Ψ constructed by placing vertices in the centers
of faces and connecting vertices if the corresponding faces of Ψ share an edge. Now
construct a subgraph Ω = (V, E, F ) of Ψ∗ by deleting all the vertices corresponding to
the faces in Q and all edges incident with those vertices. (See Figure 4.2(c) with the
yellow faces as the vertex cover.) A face of degree j of Ψ that is not in Q corresponds
to a vertex of degree
j
2
in Ω; a face of Ψ that is in Q and has degree k corresponds in
Ω to a face of degree k. Thus Ω = (V, E, F ) is a plane graph such that Ω` = Ψ and
F ` = Q, and Ω is unique up to isomorphism.
Given a 3-regular plane graph Ψ with a face-only vertex covering Q, we call the
graph obtained by the process in Lemma 4.2 the reverse leapfrog of (Ψ, Q) and denote
the graph by (Ψ, Q)−` , or simply Ψ−` if Q is understood. By Lemma 1.3, any leapfrog
graph is 3-regular and has a face-only vertex covering. Combining Lemma 1.3 with
Lemma 4.2 leads to the following characterization of leapfrog graphs:
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
(a) A bipartite plane graph
(b) The leapfrog graph Ω` is
(c) The yellow faces F ` are a
Ω = (V, E, F )
shown superimposed on Ω.
face-only vertex covering of
63
Ω` . The red and blue faces
of Ω` correspond to the vertex bipartition of Ω.
Figure 4.2: The leapfrog transformation of a bipartite plane graph
Theorem 4.3. A plane graph Ψ is a leapfrog graph if and only if Ψ is 3-regular and
has a face-only vertex covering.
Lemma 4.4. Given a bipartite plane graph Ω = (V, E, F ) with vertex partition V =
U ∪ W , the sets of faces U ` , W ` , F ` form a face 3-coloring of Ω` .
Proof. The set of faces F ` is a face-only vertex covering for Ω` . The remaining faces
in Ω` correspond to vertices in Ω. Since Ω is bipartite, each face in F ` is bounded
by alternating faces from U ` and W ` . By Lemma 1.3 any leapfrog graph is 3-regular.
Thus each vertex in Ω` is incident with exactly one face from each of the sets U ` , W `
and F ` .
Theorem 4.5. Suppose Ψ is a 3-regular plane graph. Then the following are equivalent:
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
64
1. Ψ is bipartite;
2. Ψ is face 3-colorable;
3. Ψ = Ω` for some bipartite plane graph, Ω.
Proof. (1 ⇐⇒ 2) Ψ is bipartite if and only if all faces have even degree. Thus by
Theorem 2.1, Ψ is face 3-colorable if and only if it is bipartite.
(2 =⇒ 3) The graph Ψ is 3-regular, so if Ψ has a proper face 3-coloring, each
vertex is incident with a face of each color and each color class Qi is a face-only vertex
covering. By Lemma 4.2, there is a unique graph Ωi acquired by taking the reverse
leapfrog of Ψ with respect to the vertex-covering Qi for each i. In Ωi = (Ψ, Qi )−` ,
the faces of Qi correspond to the faces of Ωi . The remaining face-only vertex covers,
Qj , Qk , correspond to the vertices of Ωi . Since Qj and Qk are independent in Ψ, the
corresponding vertices form a bipartition of the vertices in Ωi .
(3 =⇒ 2) Suppose that Ψ = Ω` for some bipartite plane graph Ω with vertex
bipartition V = U ∪ W . By Lemma 4.5, the sets of faces U ` , W ` , F ` form the color
classes for a face 3-coloring of Ψ.
4.2
Fullerenes with Complete Clar Structures
An {(a, b), k}-sphere is a k-regular plane graph with faces only of degrees a and b.
Fullerenes are exactly the class of {(5, 6), 3}-spheres. Given a fullerene Γ with a
complete Clar structure (C, A), we see that the six pairs of pentagons together with
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
65
Figure 4.3: Expansion of Γ over an edge in A
the edges between them are equivalent to the open chains discussed in Chapter 2 and
3. As in Section 3.1, we define the expansion of the fullerene as follows: widen each
of the six edges in A between pentagonal pairs into a quadrilateral faces. Each vertex
covered by A becomes an edge, and each pentagon becomes a hexagon. We denote
this new {(4, 6), 3}-sphere by E(C, A) and the set of quadrilateral faces by A0 . If Γ
has f faces, then E(C, A) has six quadrilateral faces and f hexagonal faces.
Lemma 4.6. Let Γ be a fullerene with a complete Clar structure (C, A). Let E(C, A)
be the expansion of Γ and let C 0 = C ∪ A0 . Then:
1. There is a face 3-coloring of E(C, A) such that the set C 0 forms one color class.
2. The reverse leapfrog (E(C, A), C 0 )−` = Θ is a {(4, 6), 3}-sphere.
3. The two face color classes of E(C, A) other than C 0 correspond to the vertex
color classes of the bipartite graph Θ.
Proof.
1. By Lemma 3.5, the expansion E(C, A) is face 3-colorable with the faces in C 0
forming one color class.
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
66
2. Since E(C, A) is a 3-regular graph and C 0 is a face-only vertex covering of
E(C, A), there is a unique graph Θ such that Θ` = E(C, A) and the faces of Θ
correspond to C 0 by Lemma 4.2. The faces of E(C, A) in C 0 are of degree 4 and
6, so by Lemma 1.3, the faces of Θ are of degree 4 and 6. The remaining faces
of E(C, A) are hexagons and correspond to the vertices of Θ. Thus by Lemma
1.3, Θ is 3-regular. So Θ is a {(4, 6), 3}-sphere.
3. By Theorem 4.5, the remaining two face color classes of E(C, A) correspond to
a bipartition of the vertices in Θ.
We see that a fullerene with complete Clar structure corresponds to a {(4, 6), 3}sphere Θ. We want to determine the conditions under which a {(4, 6), 3}-sphere
corresponds to one or more fullerenes with complete Clar structure.
Consider an arbitrary {(4, 6), 3}-sphere Θ and let Θ` be the leapfrog graph of Θ.
By Lemma 1.3, Θ` is also a {(4, 6), 3}-sphere; Θ` is 3-regular and has a face of degree
j corresponding to each face of degree j from Θ and a face of degree 6 corresponding
to each vertex of Θ. Also by Lemma 1.3, the faces of Θ` corresponding to the faces
from Θ form a face-only vertex covering of Θ` . Thus the quadrilateral faces of Θ`
are part of an independent set, and each quadrilateral face in Θ` is bounded by four
hexagons.
For each quadrilateral face of Θ` , choose two opposite hexagons adjacent to the
quadrilateral to form a pair. We denote this set of paired hexagons by P. We define
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
67
Figure 4.4: The reverse expansion around a quadrilateral face in (Θ` , P).
the reverse expansion procedure on the pair (Θ` , P) as follows: for each quadrilateral
face of Θ` , contract each of the two opposite edges that the quadrilateral shares with
paired hexagons into a vertex. We then obtain the graph E −1 (Θ` , P), the reverse
expansion of Θ` with respect to P. Now each quadrilateral of Θ` has become an edge
in E −1 (Θ` , P) and the degree of each face in P is decreased by 1 for each quadrilateral
that the face is paired over.
Lemma 4.7. Let Θ` be the leapfrog of a {(4, 6), 3}-sphere Θ. Let P be the set of
hexagons paired on opposite sides of the quadrilateral faces of Θ` . The reverse expansion of this pair, Γ = E −1 (Θ` , P) is a fullerene if and only if P is a set of twelve
distinct hexagons. When Γ is a fullerene, Γ admits a complete Clar structure. Γ
also admits a complete Fries structure if and only if the hexagons in P are part of a
face-only vertex covering of Θ` .
Proof. It follows from Euler’s formula that any {(4, 6), 3}-sphere has exactly six
quadrilateral faces. Thus in Θ` , there are six quadrilaterals across which we pair
opposite hexagons to create the set P. If P is a set of twelve distinct hexagons, then
each hexagon belongs to exactly one pairing and is contracted into a pentagon. Hence
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
68
Γ is a fullerene. If |P| < 12, then some hexagon in P is paired across two or more
quadrilaterals. That hexagon is contracted into a face of degree less than 5 and thus
Γ is not a fullerene.
The quadrilaterals of Θ` are part of a face-only vertex covering corresponding to
the faces of Θ by Lemma 1.3. Denote this face-only vertex covering by C 0 . Let C
denote the hexagons of Γ = E −1 (Θ` , P) corresponding to hexagons in C 0 of Θ` and let
A denote the set of edges formed by collapsed quadrilaterals from Θ` . The set C ∪ A
forms a vertex covering of Γ, and thus if Γ is a fullerene, then (C, A) is a complete
Clar structure for Γ.
Every face of Θ` has even degree, and thus Θ` is face 3-colorable by Theorem 2.1.
A face 3-coloring of Θ` corresponds to an improper face 3-coloring of Γ in which the
only incompatible faces are those that share an edge in A. Hence the pentagons of Γ
are in the same color class exactly when the faces of P are in the same color class of
Θ` , that is, if and only if P is part of a face independent set since Θ` is 3-regular. The
pentagons are all in one color class in the improper face 3-coloring if and only if they
are part of of a face-only vertex covering, which is equivalent to the condition that
the trivalent graph Γ be a leapfrog graph by Theorem 4.3. By Proposition 1.4, the
fullerene Γ is a leapfrog graph if and only if Γ admits a complete Fries structure.
When is it not possible to find a disjoint set P of opposite hexagonal faces adjacent to each quadrilateral? Consider the {(4, 6), 3}-sphere Θ. Quadrilaterals in Θ
correspond to quadrilaterals in Θ` . The four hexagons bounding a quadrilateral in Θ`
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
69
correspond to the vertices of the quadrilateral in Θ. Pairing disjoint hexagons with
quadrilaterals in Θ` is equivalent to choosing diagonal vertices for each quadrilateral
in Θ. These diagonal vertices become opposite hexagons in Θ` , and contract into
pentagons in the fullerene. We indicate these vertices around a quadrilateral face in
Θ by drawing a diagonal line connecting the vertices. (See Figure 4.5.) We define a
diagonalization of a {(4, 6), 3}-sphere Θ to be a choice of pairs of diagonal vertices
for each quadrilateral face so that no vertex is chosen twice. We can contract Θ`
into a fullerene with a complete Clar structure exactly when a diagonalization of Θ
is possible. A perfect diagonalization of a {(4, 6), 3}-sphere Θ, is a diagonalization in
which all vertices chosen are in the same cell of the bipartition and hence correspond
to faces in the same color class of a face 3-coloring of Θ` by Lemma 4.5. We have
proven the following lemma:
Lemma 4.8. Let Θ be a {(4, 6), 3}-sphere. Then
1. Θ` admits a pairing of hexagons with |P| = 12 if and only if Θ admits a diagonalization.
2. Θ` admits a pairing of hexagons with |P| = 12 where P is part of a face-only
vertex covering of Θ` if and only if Θ admits a perfect diagonalization.
Lemma 4.9. Let Θ be a {(4, 6), 3}-sphere. Then
1. Θ admits a diagonalization if and only if no vertex meets three quadrilaterals.
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
(a) A diagonal pair of vertices inci-
70
(b) Take the leapfrog, Ω` .
dent with a quadrilateral in Ω.
(c) The diagonal vertices in Ω cor-
(d) The hexagons become pen-
respond to hexagons in Ω` .
tagons in E −1 (Θ` , P).
Figure 4.5: Choose diagonal vertices for each quadrilateral in Θ. These vertices
correspond to opposite hexagons bounding a quadrilateral face in Θ` . We collapse an
edge of each of the paired hexagons to construct E −1 (Θ` , P).
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
71
2. Θ admits a perfect diagonalization if and only if no vertex in Θ is incident with
more than one quadrilateral.
Proof. There are at most three quadrilateral faces at any vertex in Θ. Since Θ is
3-regular, any two faces that share a vertex share an edge containing that vertex.
Suppose some vertex x is incident with three quadrilateral faces s1 , s2 , and s3 . One
easily checks that any multiset of diagonal edges through these quadrilaterals includes
some vertex twice. (See Figure 4.6.) If every vertex in Θ is incident with at most two
quadrilaterals, then one easily checks that we may choose diagonal vertices incident
with each quadrilateral such that no vertex is chosen twice.
If two quadrilaterals s and s0 share an edge, then the vertices incident with s
and s0 must both be part of any diagonalization, and so the diagonalization cannot
be perfect; the vertices chosen cannot be in the same cell of the bipartition. If the
quadrilaterals are disjoint and U is one cell of the vertex bipartition, we may choose
diagonal vertices in U for each quadrilateral.
Theorem 4.10.
1. The fullerenes on n vertices that admit a complete Clar structure are in one-toone correspondence with the diagonalized {(4, 6), 3}-spheres on
n
3
+ 4 vertices.
2. The fullerenes on n vertices that admit a complete Clar structure and a complete
Fries structure are in one-to-one correspondence with the perfectly diagonalized
{(4, 6), 3} spheres on
n
3
+ 4 vertices.
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
72
s3
s2
s1
s2
sk
s1
Figure 4.6: If three quadrilaterals in Θ share a vertex, then any choice of diagonal
vertices for each quadrilateral includes some vertex twice. If at most two quadrilaterals share each vertex, we can choose diagonal vertices for each quadrilateral of Θ
so that no vertex is chosen twice.
Proof. By Lemma 4.7, a leapfrog {(4, 6), 3}-sphere Θ` can be contracted into a fullerene
Γ with complete Clar structure exactly when it is possible to find a set of pairings
P of twelve distinct hexagons bounding opposite sides of the six quadrilateral faces
and by Lemma 4.8 this is equivalent to a diagonalization of Θ. By Proposition 1.4
and Lemma 4.7, the contracted fullerene Γ is leapfrog exactly when the hexagons we
choose to collapse are all in the same color class. By Lemmas 4.5 and 4.9, this occurs
for each perfect diagonalization of a {(4, 6), 3}-sphere.
If the fullerene Γ = E −1 (Θ` , P) has n vertices, then Θ` has n + 12 vertices; the
reverse expansion contracts the six quadrilateral faces into six edges. Thus Θ is
one-third the order of Θ` and has
n
3
+ 4 vertices.
For each {(4, 6), 3}-sphere in which no vertex lies on two quadrilaterals, two perfect
diagonalizations are possible. We have two choices for the diagonal in a quadrilat-
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
73
Figure 4.7: Two choices for contracting a quadrilateral face in Θ` . A different choice
is equivalent to a Stone-Wales transformation of the fullerene.
eral face. For each of the other quadrilaterals, we must choose diagonal vertices in
this same bipartition. Thus for each {(4, 6), 3}-sphere on which no two quadrilaterals
share a vertex, there are two perfectly diagonalized {(4, 6), 3}-spheres and two corresponding fullerenes with a complete Fries structure and a complete Clar structure.
For each diagonalization of a {(4, 6), 3}-sphere Θ, Θ` can be contracted into a
different fullerene with complete Clar structure. How many such diagonalizations are
there for a {(4, 6), 3}-sphere? Given a quadrilateral face in Θ, we have two choices
for the pair of diagonal vertices incident with that face. Suppose two quadrilaterals,
s1 and s2 , share a vertex in Θ. Then s1 and s2 share an edge, and a diagonal pair of
vertices incident with s1 includes one vertex on that edge. A diagonal pair of vertices
incident with s2 must also include a vertex on this edge, and there is only one choice
remaining for the pair of diagonal vertices incident with s2 .
We call a set of quadrilaterals in Θ a block if each of the quadrilaterals in the
set shares a vertex with another quadrilateral in the set. Once we choose a pair of
diagonal vertices for one quadrilateral, the choice of diagonal vertices is forced for each
of the other quadrilaterals in the block. (See Figure 4.6.) Thus if Θ has b blocks, there
CHAPTER 4. FULLERENES WITH COMPLETE CLAR STRUCTURE
74
are 2b diagonalized {(4, 6), 3}-spheres and 2b corresponding fullerenes with complete
Clar structure. Note that 1 ≤ b ≤ 6, so a given {(4, 6), 3}-sphere could yield as many
as 64 fullerenes with complete Clar structure. Choosing a different diagonal in Θ
corresponds to choosing the other pair of hexagons around a quadrilateral in Θ` to be
in P. Choosing a different pair of hexagons around a quadrilateral in Θ` to contract
is equivalent to a Stone-Wales transformation [4] of the four faces in the fullerene (see
Figure 4.7). We can get from any one of these fullerenes to another through a series
of Stone-Wales transformations. Thus each {(4, 6), 3}-sphere Θ in which no vertex
lies on three quadrilaterals represents an equivalence class of fullerenes with complete
Clar structure under the Stone-Wales equivalence.
75
Chapter 5
Class for which the Clar and Fries
Number Cannot be Attained by
the Same Kekulé Structure
5.1
Introduction
It is part of the folklore of fullerenes that a set of independent benzene faces that
attains the Clar number for a fullerene is contained in the set of benzene faces that
gives the Fries number. In this chapter, we describe a class of fullerenes for which
this does not hold: for fullerenes in this class, any Kekulé structure that attains the
Fries number cannot give the Clar number; any Kekulé structure that attains the
Clar number cannot give the Fries number.
CHAPTER 5. CLAR AND FRIES CLASS
76
A face 3-coloring is not possible over a fullerene, but a partial 3-coloring can be
constructed except over relatively small excluded patches containing the pentagonal
faces. We can begin to construct a Kekulé structure consisting of edges connecting
the faces in one color class as described previously. This structure must be extended
inside each excluded patch to complete the Kekulé structure. To attain the Fries
number, this extension must be chosen so that |B1 (K)| + 2|B2 (K)| is minimized. To
achieve the Clar number, this extension must be chosen so that |A|, the number of
Kekulé edges not on a Clar face, is minimized. In the examples we construct here,
pairs of nonadjacent pentagons are joined by a single edge, and we refer to such
patches as basic patches. These basic patches are widely separated to ensure that no
other pairing of pentagons could yield the Fries or Clar number.
To construct the partial Kekulé structure outside of the basic patches, we choose
one color class of independent faces to be the void faces. We must then choose another
color class to be the set C contributing to the Clar number. Thus there are six possible
options for choosing the partial Kekulé structure and the partial Clar structure. In
Section 5.2, we show that for exactly one of these six choices around a basic patch, no
completion of the Kekulé structure simultaneously minimizes the contribution to |A|
and |B1 (K)| + 2|B2 (K)| over the patch. In Section 5.3, we construct fullerenes with
six basic patches so that for each of the six choices for void and Clar faces, exactly
one of these basic patches requires different extensions of the Kekulé structure to
minimize the Clar deficit and the Fries deficit. Thus for our class of fullerenes, no
CHAPTER 5. CLAR AND FRIES CLASS
77
C
V
(a) A partial face 3-coloring out-
(b) A partial Kekulé structure
side of a basic patch.
given by the choice of void faces
outside of the basic patch.
Figure 5.1: The white faces represent a basic patch. The void faces are contained in
the set of blue faces, the Clar faces in the set of pink faces.
Kekulé structure attains both the Fries and the Clar number for the fullerene.
5.2
5.2.1
Basic Patches
A choice for the void and Clar faces that requires two
Kekulé extensions.
Figure 5.1(a) depicts a basic excluded patch. For the faces surrounding the basic
patch, the set of blue faces is chosen to be the set of void faces and the pink faces
are chosen to be Clar faces. Figure 5.1(b) shows the partial Kekulé structure given
CHAPTER 5. CLAR AND FRIES CLASS
78
by this choice of void faces; all edges that join two blue faces are in the partial
Kekulé structure. We need to extend this to a Kekulé structure. Figure 5.2(a) shows
the extension of this choice that minimizes |A| and 5.2(b) shows the extension that
minimizes |B1 (K)| + 2|B2 (K)|. We prove that no single extension minimizes both
deficits.
Extension 1: We first extend the Kekulé structure to minimize the Clar deficit.
Outside the patch, we start with a partial Kekulé structure in which each pink hexagonal face is a benzene face (Figure 5.1(b)). None of the ten vertices incident with
one of the two pentagons is covered by a face in C, so these vertices must be covered
by edges from A in the vertex covering (C, A). In the partial Kekulé structure, every
hexagon that is not in C is adjacent to a face in C. Thus no extension can increase
|C| over the patch. Any extension that does not reduce |C| must cover only the ten
vertices incident with the pentagons. There is only one perfect matching for these
ten vertices, and it is shown as a completion of the Kekulé structure in Figure 5.2(a).
Note that this is a Clar chain of Type 3 between the two pentagons. Over the patch in
this extension, |A| = 5 and |B2 (K)| = 6, |B1 (K)| = 4, giving |B1 (K)|+2|B2 (K)| = 16
Extension 2: The Kekulé structure in Figure 5.2(b) has |A| = 8 and |B2 (K)| = 4,
|B1 (K)| = 2, giving |B1 (K)|+2|B2 (K)| = 10. While |A| is not minimized, B2 (K) and
B1 (K) are both smaller than in Extension 1. Extension 1 is the only extension that
minimizes the Clar deficit, and that extension does not minimize the Fries deficit.
Thus for this choice of void and Clar faces over a basic patch, any structure that
CHAPTER 5. CLAR AND FRIES CLASS
(a) Extension 1 minimizes the Clar
(b) Extension 2 minimizes the Fries
deficit. Here |A| = 5 and |B2 (K)| =
deficit. Here |A| = 8 and |B2 (K)| =
6, |B1 (K)| = 4.
4, |B1 (K)| = 2.
79
Figure 5.2: Extension 1 and Extension 2 on a basic patch where the Clar faces are
pink and the void faces are blue.
contributes the maximum number of faces toward the Clar number over this patch
cannot achieve the maximum number of benzene faces.
5.2.2
Extending Kekulé structures over basic patches with
other choices for the void and Clar faces
We show that the choice for the void faces and faces in C described in Section 5.2.1
is the only case over such a patch for which |A| and |B1 (K)| + 2|B2 (K)| cannot be
CHAPTER 5. CLAR AND FRIES CLASS
80
minimized simultaneously. There are five options remaining for the choice of void
and Clar faces around the patch, and for each of these choices, only one extension is
necessary to minimize both the Clar and Fries deficits.
Suppose we let the blue faces be the void faces and the yellow faces be the Clar
faces. Then the Kekulé structure in Figure 5.2(b) has a minimal Fries deficit of
|B1 (K)| + 2|B2 (K)| = 10. Every yellow hexagon is a benzene face, so we also have a
maximum number of faces contributing to the Clar count, with |A| = 2 (a Clar chain
of Type 2).
Suppose that the void faces are in color class that includes the pentagons (here,
the pink faces). Then the edges joining these faces complete a Kekulé structure over
the patch, as seen in Figure 5.3(b). |B1 | = |B2 | = 0, so the number of benzene faces
over the patch is clearly maximized. We must also choose a color class to be the Clar
faces. Since all hexagons in the remaining two color classes are benzene faces, |C|
is also maximized over the patch for either choice. Thus the same Kekulé structure
maximizes the number of Clar faces and the number of benzene faces over the patch.
Suppose that the yellow faces are the void faces. Begin a partial Kekulé structure
consisting of all edges that join two yellow faces. Extend this Kekulé structure so that
all blue and pink hexagons are benzene faces as in Figure 5.3(c). For either choice
of the Clar faces, |C| is clearly maximized over this patch. |B1 | = |B2 | = 2, and any
local change increases |B1 (K)| + 2|B2 (K)| and decrease the number of benzene faces.
Thus both the Clar and Fries deficits are minimized in this extension.
CHAPTER 5. CLAR AND FRIES CLASS
81
(a) A basic patch with a partial
(b) The pink faces represent the
face 3-coloring.
void faces.
(c) The Yellow faces are void.
Figure 5.3: Over a basic patch, we choose other color classes for the void and Clar
faces and consider extensions of the resulting Kekulé structure.
CHAPTER 5. CLAR AND FRIES CLASS
82
2
1.5
1.5
1.5
(1,1)
2
(r+s)
(r)
r ≠ 0 mod 3
r = s mod 3
Figure 5.4: The vertices of this auxiliary graph represent the pentagons in a fullerene.
The edges give the Coxeter coordinates of the segments between nearby pentagons.
We see that for every case except that described in section 5.2.1, the same Kekulé
structure maximizes the number of faces contributing to the Clar number and the
Fries number over the basic excluded patch.
5.3
Fullerenes over which the Clar and Fries numbers cannot be attained simultaneously
We saw in the previous section that there are six choices for the void faces and the Clar
faces. We also saw that given an excluded patch, the Kekulé structure of all but one
of these choices can be extended to the basic patch while simultaneously maximizing
the number benzene faces and the number of Clar faces. To force the existence of a
CHAPTER 5. CLAR AND FRIES CLASS
83
patch in which these parameters cannot be maximized by the same Kekulé structure,
we need a fullerene with patches so that one of them is the exceptional patch for each
the six choices. One infinite class of examples can be found in Graver’s “Catalog of
Highly Symmetric Fullerenes,” pictured in Figure 5.4 with segment and angle types.
The paper describes these parameters [6]. For our purpose, it is only necessary to
understand that the vertices represent pentagonal faces and in our case the green
edges represent the excluded patches which, in our example, are pairs of pentagons
joined by a single edge. Furthermore, the partial face 3-coloring is different around
each of the six excluded patches. Thus regardless of which of the six color choices
for the color class of the void faces and the color class of the Clar faces is made,
one of the six patches is such that one Kekulé structure maximizes the Clar number,
while another Kekulé structure maximizes the Fries number, and the two parameters
cannot be maximized simultaneously.
An example with s = 1, r = 7 is shown on the next page. In the next chapter,
we show that a fullerene in this class with r ≥ 7 is widely separated. Thus the
Clar number is achieved by a Kekulé structure with Clar chains between pentagons
together in basic patches. In this coloring, the red faces indicate the set containing
C and the void set is contained in the blue color class. A pair of pentagons lies in
a basic patch on each interior corner. The edges in A are represented by thick red
edges and the remaining edges in the Kekulé structure are represented by thick blue
edges. A blue arrow indicates the excluded patch over which the number of faces in
CHAPTER 5. CLAR AND FRIES CLASS
84
C and the number of benzene faces cannot be maximized simultaneously.
In the first figure, there are 272 benzene faces and there are 135 faces in C. In the
second figure, there are 274 benzene faces and 134 faces in C. The first figure attains
the Clar number but not the Fries number for the fullerene, and the reverse is true
for the second. Hence the set of faces that attains the Clar number is not contained
in a set of faces that attains the Fries number.
CHAPTER 5. CLAR AND FRIES CLASS
85
2
a.
1
1
2
2
1
2
2
1
1
2
2
1
2
2
2
1
1
1
2
2
2
1
2
2
1
1
2
b.
2
1
2
1
2
1
2
1
2
2
2
1
1
2
2
2
Figure 5.5: The Clar faces are red and the void faces are blue. An arrow indicates
the patch over which the Fries and Clar deficits cannot be minimized simultaneously.
Figure (a) minimizes the Clar deficit, Figure (b) minimizes the Fries deficit. The
numerals represent faces in the sets B1 and B2 .
86
Chapter 6
Examples and Future Research
In this Chapter, we use the methods established in Chapters 2 and 3 to find the
Clar number for several classes of fullerenes. The first two examples are taken from
Graver’s “A Catalog of All Fullerenes with Ten or More Symmetries” [6], and the Clar
number is found when the parameters are such that we have pairs of pentagons that
are widely separated. In the next section, we find the Clar number for Icosahedral
Leapfrog fullerenes. Recall from Lemma 3.3 that for a fullerene with a Clar structure
(C, A), the number of faces in C is
|V |
6
−
|A|
.
3
We regularly use Lemmas 3.13, 3.14,
and 3.15, which state that for a pair of pentagons joined by a segment with Coxeter
coordinates (m, n) where m ≥ n, a chain of Type 1 contributes m edges to A, a chain
of Type 2 contributes m + n edges to A, and a chain of Type 3 contributes 3m + 2
edges to A.
CHAPTER 6. EXAMPLES AND FUTURE RESEARCH
a
b
to a
1.5
1.5
1.5
(p,p)
2
(r+s)
c
f
87
to b'
a'
(r)
to f'
e
d
(a) The vertices of this auxiliary graph repre-
(b) The pentagon a0 in the fullerene Γ is
sent the pentagons in a fullerene. The edges
represented by a red vertex of the auxil-
give the Coxeter coordinates of the segments be-
iary graph. Here we see the patch sur-
tween nearby pentagons.
rounding the pentagon a0 .
Figure 6.1: Class of Fullerenes that generalizes the family described in Chapter 5.
The pentagons are paired over green segments with Coxeter coordinates (p, p).
6.1
Two Classes of Widely Separated Fullerenes
We first find the Clar number for a family of fullerenes that generalizes the class
given in Chapter 5. Figure 6.1(a) shows an auxiliary graph that represents a general
fullerene in this class. The vertices of the auxiliary graph represent the twelve pentagons in the fullerene. The edges represent segments between nearby pentagons, and
the colors code the Coxeter coordinates of these segments, defined by the parameters
p, r and s. The numbers shown in Figure 6.1(a) represent angle types between two
segments joined by a common pentagon, and the meaning of these numerals is shown
in Figure 6.1(b). For a detailed description, see [6]. Different choices for parameters
r, p, and s result in all fullerenes within this family. Graver showed in [6] that the
CHAPTER 6. EXAMPLES AND FUTURE RESEARCH
88
number of vertices for a fullerene in this family is 12r2 + 2s2 + 12rs + 12p(2r + s).
These fullerenes are widely separated when p is much smaller than r (an inequality is
given shortly). In this case, the six open Clar chains must pair pentagons joined by
segments with Coxeter coordinates (p, p), represented by green edges in Figure 6.1.
There are several cases depending on the congruence classes of r and s modulo 3. We
consider the congruence in Chapter 5 as well as the congruence resulting in a leapfrog
fullerene.
Suppose that r 6≡ 0 (mod 3) and r ≡ s (mod 3), giving r + s 6≡ 0 (mod 3). Let
a, b, c, d, e, f be pentagonal faces on the fullerene in clockwise order as shown in Figure
6.1. In a partial face 3-coloring that avoids the Clar chains between segments with
Coxeter coordinates (p, p), a and b are in different color classes since the segment
between a and b has coordinate (r + s), where r + s 6≡ 0 (mod 3). Similarly, the
segment between b and c has Coxeter coordinate (r), and so b and c are in different
color classes. Since r 6≡ 0 but r ≡ s (mod 3), r + s 6≡ r (mod 3). Thus a and c are
in different color classes. We see that a and d are in one color class, b and e are in a
second color class, and c and f are in a third color class. Each of these pentagons is
paired over a segment with another pentagon and the Coxeter coordinates over these
segments are (p, p), so each pair is in the same color class. We want to compare the
segment types for faces in the same color class. Without loss of generality, say that
a and d are red, b and e are yellow, and c and f are blue. Note that the Coxeter
coordinates between a and the yellow face b are (r + s), the coordinate between d and
CHAPTER 6. EXAMPLES AND FUTURE RESEARCH
89
the yellow face e is (r). There are two possibilities for the position of the yellow color
class around a red pentagon. The pentagon a and the pentagon d must each have
a different position with respect to the yellow faces. Thus all six possible colorings
around these segments are represented. The type of Clar chain only depends on
the position of the faces in the color class containing C. This class is symmetric,
and regardless of the color chosen, there are two chains of each of the three types.
Thus the total contribution to A is 2p + 2(p + p) + 2(3p + 2) = 12p + 4. The next
closest pentagons that are unpaired have coordinates (r). This choice of Clar chains
is widely separated when r ≥
12p+4
2
− 2 = 6p. For this class, the number of vertices is
|V | = 12r2 + 2s2 + 12rs + 12p(2r + s). Thus, when the chains are widely separated,
the Clar number is
12r2 + 2s2 + 12rs + 12p(2r + s) 12p + 4
−
6
3
1
= 2r2 + (s2 − 4) + 2rs + 4rp + 2p(s − 2)
3
Note that s 6≡ 0 (mod 3), and so s2 ≡ 1 (mod 3). Thus the above expression is
always an integer.
Suppose that r ≡ s ≡ 0 (mod 3). By Proposition 1.4, this is a leapfrog fullerene.
All of the pentagons are in the same color class, say the set of red faces. We choose
one of the remaining color classes to contain the set C in order to avoid having any
chains of Type 3. Consider one pair of pentagons with Coxeter Coordinates (p, p)
and choose C to be the color class that allows this chain to be of Type 1. This
CHAPTER 6. EXAMPLES AND FUTURE RESEARCH
90
is possible because the chain type is defined by the position of the Clar faces with
respect to the segment between the two pentagons. Once we choose the set C around
the first segment, the types for the remaining segments are forced. As we go around
this chain, the red color class remains the same (it is a perfect face-independent set)
and the remaining two colors are transposed. Thus, at the next pair of pentagonal
faces, the position of the remaining two color classes in relation to the pentagons has
switched, and the segment is of Type 2. The segment types alternate between Type
1 and Type 2, so we have three chains of Type 1, each contributing p edges to A,
and three chains of Type 2, each contributing 2p edges to A. The total number of
edges in A is 9p. Again, the Coxeter coordinates between the next closest unpaired
pentagons is (r). The (p, p) Clar chains are widely separated when r ≥
9p
2
− 2. In
this case, the Clar number is
12r2 + 2s2 + 12rs + 12p(2r + s) 9p
−
6
3
1
= 2r2 + s2 + 2rs + 4rp + 2ps − 3p
3
Since s ≡ 0 (mod 3), the expression is always an integer.
We compute the Clar number for another class of fullerenes with a different symmetry group, and the auxiliary graph for this class is pictured in Figure 6.2. Again,
the vertices of the auxiliary graph represent pentagons. The five segments with arrows connect to a common pentagon. The fullerenes in this class are attained by
choosing values for the parameters r and s. Graver showed in [6] that the number of
vertices for a fullerene in this class is 24p2 + 48pr + 20r2 . The fullerenes are widely
CHAPTER 6. EXAMPLES AND FUTURE RESEARCH
91
(p+r, p)
(p, p+r)
(r)
Figure 6.2: The auxiliary graph for a class of symmetric fullerenes. Vertices represent pentagons, edges show the Coxeter coordinates of segments between nearby
pentagons. The pentagons paired over segments with coordinates (r).
separated when r ≡ 0 (mod 3) and r is small in comparison with p (an inequality
follows shortly). The paired pentagons are then connected by segments with Coxeter
coordinates (r), and the six open Clar chains are between these pairs, represented by
green edges in Figure 6.2. All coordinates between nearby pentagons are congruent
modulo 3, so this is a leapfrog fullerene by Proposition 1.4. All of the pentagonal
faces are in the same color class, so we choose the set C to be either of the two
independent sets of faces that do not contain the pentagons. Then each of the Clar
CHAPTER 6. EXAMPLES AND FUTURE RESEARCH
92
chains must be of Type 1 or Type 2. For segments with only coordinate (r), Type 1
and Type 2 each contribute r edges to A. Thus the total number of edges in A over
the fullerene is 6r. We see that these chains are widely separated if for any other
pair of pentagons, one of the Coxeter coordinates is at least
6r
2
− 2 = 3r − 2. The
next closest pairs have coordinates (p + r, p) and (p, p + r). Thus if p ≥ 2r − 2, then
chains with Coxeter coordinates (r) are widely separated. The number of vertices for
fullerenes in this class is |V | = 24p2 + 48pr + 20r2 , so the Clar number is
|V | |A|
24p2 + 48pr + 20r2 6r
10
−
=
−
= 4p2 + r2 + 8pr − 2r.
6
3
6
3
3
The restriction that r ≡ 0 (mod 3) ensures that this is always an integer.
These few examples were chosen to illustrate our computational approach to the
Clar number. Using these techniques in conjunction with the Catalog [6], the Clar
number can be easily computed for many infinite families of fullerenes. In the next
section, we employ our theory to compute the Clar number for a family of fullerenes
in which the pentagons are not widely separated.
6.2
Icosahedral Leapfrog Fullerenes
The natural generalization of C60 is the class of icosahedral leapfrog fullerenes. To
construct an icosahedral fullerene, choose an equilateral triangle from the hexagonal
tessellation with vertices at the centers of hexagons and copy this triangle onto each
face of the icosahedron. The result is a fullerene with the twelve pentagonal faces at
CHAPTER 6. EXAMPLES AND FUTURE RESEARCH
C60
93
(4,2)
Icosahedral
Fullerene
Dodecahedron
Figure 6.3: To form an icosahedral fullerene, replace each face of an icosahedron with
an equilateral triangle from the hexagonal tessellation with vertices at the centers of
faces.
the twelve vertices of the icosahedron. This construction was first given by Coxeter
[2]. An icosahedral fullerene is uniquely determined by the Coxeter coordinates (m, n)
of the sides of the triangle, and the icosahedral fullerene is a leapfrog fullerene exactly
when m and n are congruent modulo 3.
Consider an icosahedral leapfrog fullerene with parameters (m, n) and assume
without loss of generality that m ≥ n. Since the fullerene is leapfrog, any pair
of pentagons can be connected by a Clar Chain. The shortest segments between
pentagons have Coxeter coordinates (m, n), and thus 6m is a lower bound for A,
which would be achieved if we could pair pentagons with six Clar chains of Type 1.
Attaining this lower bound for |A| would show that the Clar number for Γ is
We show that such a pairing is possible.
|V |
6
− 2m.
CHAPTER 6. EXAMPLES AND FUTURE RESEARCH
94
P2
H2
P4
P1 H1
H3
P3
P4
P6
P5
Figure 6.4: Five equilateral triangles that share P3 as a vertex.
Theorem 6.1. Suppose Γ is an icosahedral fullerene with parameters (m, n) where
m ≥ n and m ≡ n (mod 3). Then the Clar number for Γ is |C| =
|V |
6
− 2m.
Proof. Since all of the segments between nearby pentagons are congruent modulo 3,
the pentagons are all in one color class, say red, in any an improper face 3-coloring
derived from a Clar structure. To attain the Clar number, we must choose the color
class containing C to be one of the remaining two colors. Consequently, all chains are
of Type 1 or Type 2. Choose a pair of pentagons P1 and P2 with Coxeter coordinates
CHAPTER 6. EXAMPLES AND FUTURE RESEARCH
95
(m, n), to connect with a Clar chain. We can then choose the color class for C so that
the chain is of Type 1. We may assume that this is the blue color class, and begin
a partial face 3-coloring. Now choose a pentagon P3 that completes the equilateral
triangle with P1 and P2 . There are five equilateral triangles with a P3 as a vertex,
with edges {P1 , P2 }, {P2 , P4 }, {P4 , P5 }, {P5 , P6 }, {P6 , P1 } in clockwise order (see
Figure 6.4). The improper face 3-coloring in Figure 6.4 results when the Clar chain
connecting P1 and P2 is of Type 1 and forces the chain connecting P3 and P5 to also
be of Type 1. We now show that this holds in general.
Assume that n 6= 0. (If all the segments have coordinates (m), then Type 1 and
Type 2 are the same, and any pairing will work). There is an (m, n) segment from
P3 to P1 ; consider the (m − 1, n − 1) segment contained in this segment between
hexagons H3 and H1 adjacent to P3 and P1 , respectively. The coordinates between
these two faces are also congruent modulo 3, so these faces are in the remaining color
class, say yellow. We have constructed a Clar chain of Type 1 between P1 and P2 ,
and the face H1 is incident with an edge of this Clar chain. The faces of the partial
3-coloring alternate yellow and blue around the red pentagon P3 . Thus as we reach
the second face from H3 on either side of the pentagon, both are in the yellow color
class, and shares an edge of the Clar chain. This is a Clar chain of Type 1 between
the pentagons P3 and P5 .
If we choose Clar chains between adjacent pairs of pentagonal faces so that all
nearby segments have this relationship, then each of the Clar chains can be of Type
CHAPTER 6. EXAMPLES AND FUTURE RESEARCH
96
7
7
12
5
6
6
1
7
2
1
12
7
4
3
11
9
11
8
9
7
10
10
11
7
7
7
Figure 6.5: The Leapfrog Icosahedron with a pairing of pentagons, all of Type 1.
1 with length m. There are many such pairings, and one is shown for an arbitrary
icosahedral fullerene in Figure 6.5. This set of chains meets the lower bound |A| = 6m
over the icosahedral fullerene, and so the Clar number is of Γ is |C| =
6.3
|V |
6
− 2m.
Future Research
A major area for future research is the general structure of chain decompositions.
In particular, what is the Fries analog to Clar chains? We now understand noninterfering Clar chains, and can use them to find the Clar number for fullerenes with
widely separated pairs of pentagons. The interaction between two or more chains is
not understood; if chains share adjacent faces, the number of edges contributed to
CHAPTER 6. EXAMPLES AND FUTURE RESEARCH
97
A is often different from when the chains are non-interfering. As we saw in Figure
3.4, chains may also zigzag around pairs of pentagons, and the ability to connect two
pentagons through a Clar chain may depend upon the relation of the chain to other
pentagons.
As noted above, we can use Clar chains to find the Clar number for many classes of
fullerenes with widely separated pairs of pentagons. This method could be applied to
other classes of fullerenes to catalog the Clar number for large classes of fullerenes. We
would like to have an analogous understanding of widely separated sets of four or six
pentagons (we need only consider even groupings of pentagons, because we know that
pairs must be connected by open Clar chains). Fullerenes with two widely separated
sextets of pentagons would be a subset of the class of nanotubes. Nanotubes are
fullerenes with two caps each containing six pentagons and separated by a cylinder of
hexagons. We would like to distinguish between classes in which chains must connect
pentagons in different caps and those for which each of the open chains can connect
two pentagons within the same cap.
For fullerenes with widely separated pairs of pentagons, we have shown that any
Clar structure that attains the Clar number does not include closed Clar chains other
than those equivalent to chains of Type 3. It remains to be determined whether closed
chains may exist more generally.
We know that there are fullerenes for which the Clar number and the Fries number cannot be attained by the same Kekulé structure; that is, there is no Kekulé
CHAPTER 6. EXAMPLES AND FUTURE RESEARCH
98
structure K such that a set of faces B3 (K) attaining the Fries number contains an
independent subset attaining the Clar number. One open question would be to classify fullerenes that do satisfy this property. For example, it may be the case that in
leapfrog fullerenes, the Fries and Clar number can always be attained from the same
Kekulé structure.
In Chapter 2, we introduced chain decompositions and show that Clar and Fries
schemes can be completed into Kekulé structures for decompositions with detached
chains. We would like to show that these decompositions can be completed in general.
The decomposition of a fullerene could also be used to find bounds for the Fries
number. Under what restrictions does a chain decomposition result in the Fries
number, and when does the same decomposition result in both the Clar number and
Fries number?
Chain decompositions allowed us to find an improper face 3-coloring for fullerenes.
This approach may lead to a similar result for planar graphs in general. A 3-regular
plane graph contains open chains connecting pairs of faces of odd degree, and we can
consider an improper face 3-coloring from the expansion of the edges of these chains
as in Theorem 2.6.
99
Bibliography
[1] E. Clar, The Aromatic Sextet, Wiley, London, 1972.
[2] H.S.M. Coxeter, Virus Macromolecules and Geodesic Domes, A Spectrum of
Mathematics, J.C. Butcher, ed., Oxford Univ. Press (1971), pp. 98-107.
[3] P. W. Fowler, Localized Models and Leapfrog Structures of Fullerenes, J. Chem.
Soc., Perkin 2, (1992), pp. 145-146.
[4] P. W. Fowler and D. E. Manolopoulos, An Atlas of Fullerenes, Clarendon Press,
Oxford, 1995.
[5] P. W. Fowler and Tomaz̆ Pisanski, Leapfrog Transformations and Polyhedra of
Clar Type, J. Chem. Soc., Faraday Trans., 90 (1994), pp. 2865-2871.
[6] J. E. Graver, A Catalog of All Fullerenes with Ten or More Symmetries,
DIMACS Series in Discrete Mathematics and Theoretical Computer Science,
Vol.69, AMS, (2005), pp 167-188.
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BIOGRAPHICAL DATA
NAME OF AUTHOR: Elizabeth Hartung
PLACE OF BIRTH: Bloomsburg, Pennsylvania
DATE OF BIRTH: 26 November 1983
GRADUATE AND UNDERGRADUATE SCHOOLS ATTENDED:
• Syracuse University
• Indiana University of Pennsylvania
DEGREES AWARDED:
• M.S. 2008, Syracuse University
• B.S. 2006, Indiana University of Pennsylvania